es

Limit theorems

Recall that a null set is a Lebesgue measurable set of real numbers that has measure zero. This can be characterized as a set that can be covered by a countable union of intervals of arbitrarily small total length.
A sequence of functions { f ₙ(x) } all having the same domain X and codomain Y is said to converge pointwise to a given function f : X → Y often written as
\[ \lim_{n\to \infty} f_n (x) = f(x) \qquad\mbox{or just} \qquad \lim_{n\to \infty} f_n = f \]
if (and only if) the limit of the sequence f ₙ(x) evaluated at each point x in the domain of f is equal to f(x). The function f is said to be the pointwise limit function of the { f ₙ(x) }.

A sequence { f ₙ(x) } of extended real-valued functions on a measure space X is said to converge almost everywhere (a.e.) to the function f if there is a set E ⊂ X of points where sequence { f ₙ } does not converge pointwise to f has zero measure, i.e., μ(E) = 0.
Similarly, we say that the sequence { f ₙ } is Cauchy a.e. if there exists a set E of measure zero so that fn(x) is a Cauchy sequence of real numbers for all x not belonging to E. That is, given x ∉ E and ε > 0 there is some natural number N depending on x and ε so that whenever m,n ≥ N, we have | fn(x) − fm(x)| < ε.
This definition can be extended for a general case.
We say that a property about real numbers x holds almost everywhere (with respect to Lebesgue measure µ) if the set of x where it fails to be true has µ measure 0.
Theorem 1: Let f : ℝ ↦ [−∞, ∞] is integrable, then f(x) ∈ ℝ holds almost everywhere (or, equivalently, |f(x)| < ∞ almost everywhere).
Let E = { x : |f(x)| = ∞ }. Then we need to show that μ(E) = 0.

It is given that \( \displaystyle \quad \int_{\mathbb{R}} |f(x)|\,{\text d}\mu < \infty . \quad \) So for any integer n ∈ ℕ, the simple function nχE satisfies nχE(x) ≤ |f(x)| always and hence \[ \int_{\mathbb{R}} n\,\chi_E {\text d}\mu = n\,\mu (E) \le \int_{\mathbb{R}} |f| \,{\text d}\mu < \infty . \] However, this can’t be true for all n ∈ ℕ unless μ(E) = 0.

   
Example 1: We start with not Lebesgue integrable function \[ f(x) = \begin{cases} \infty & \text{if } x \in \mathbb{Q} \cap [0,1] \\ 0 & \text{otherwise} \end{cases} \] However, we can modify this idea to construct a function that is integrable and still takes infinite values on a countable set. Let \( \displaystyle \quad \{q_n\}_{n=1}^\infty \quad \} be an enumeration of the rational numbers in [0,1]. Define: \[ f(x) = \begin{cases} n & \text{if } x = q_n \\ 0 & \text{otherwise} \end{cases} \] This function is:
  • Infinite at each rational q_n, since f(q_n) = n \to \infty as n \to \infty
  • Zero on all irrational numbers in [0,1], which form a set of full measure
Now compute the Lebesgue integral: \[ \int_0^1 f(x) \, dx = \sum_{n=1}^\infty f(q_n) \cdot \mu(\{q_n\}) = \sum_{n=1}^\infty n \cdot 0 = 0 \] So f is Lebesgue integrable (integral is finite), even though it takes arbitrarily large (even unbounded) values on a countable set.

Let us define measurable but not integrable function f : (0, 1] → [1, ∞) by \[ f(x) = \frac{1}{x} , \qquad x \in [0, 1] . \] This function is:

  • Measurable on (0,1]
  • Finite for all x ∈ (0,1] — so it is finite everywhere on its domain
  • But not integrable over (0,1], because: \[ \int_0^1 \frac{1}{x} \, {\text d}x = \infty . \]
So f ∉ 𝔏¹((0,1]), even though f(x) < ∞ for all x ∈ (0,1]. Therefore, Theorem 1 only says: if f is integrable, then it must be finite almost everywhere — not the other way around.    ■
End of Example 1
Theorem 2: If f : ℝ ↦ [−∞, ∞] is measurable, then f satisfies \[ \int_{\mathbb{R}} |f|\,{\text d}\mu = 0 \] if and only if f(x) = 0 almost everywhere.
Suppose \( \displaystyle \quad \int_{\mathbb{R}} |f(x)|\,{\text d}\mu = 0 . \quad \) Let En = { x : |f(x)| ≥ 1/n }. Then \[ \frac{1}{n}\,\chi_{E_n} \le |f| \] and so \[ \int_{\mathbb{R}} \frac{1}{n}\,\chi_{E_n} {\text d}\mu = \frac{1}{n}\,\mu (E_n ) \le \int_{\mathbb{R}} |f|\,{\text d}\mu = 0 . \] Thus, μ(En) = 0 for each n. However, E₁ ⊆ E₂ ⊆ ⋯ and \[ \cup_{n=1}^{\infty} E_n = \left\{ x \in \mathbb{R}\, : \ f(x) \ne 0 \right\} . \] Therefore, \[ \mu \left( \left\{ x \in \mathbf{R} \ : \ f(x) \ne 0 \right\} \right) = \mu \left( \cup_{n=1}^{\infty} E_n \right) = \lim_{n\to\infty} \mu \left( E_n \right) = 0 . \]

Conversely, suppose now that    μ({ x ∈ ℝ : f(x) ≠ 0}) = 0. It is clear that |f| is non-negative measurable function and so there is a monotone increasing sequence { fn } of measurable simple functions that converges pointwise to |f|. From inequality 0 ≤ fn(x) ≤ |f| it follows that    { x ∈ ℝ : fn(x) ≠ 0 } ⊆ { x ∈ ℝ : f(x) ≠ 0 } and so    μ({ x ∈ ℝ : fn(x) ≠ 0 }) ≤ μ({ x ∈ ℝ : f(x) ≠ 0 }) = 0. Being a simple function fn has a largest value yn, which is finite, and so if we set En = { x ∈ ℝ : fn(x) ≠ 0 }, we get \[ f_n \le y_n \chi_{E_n} \quad \Longrightarrow \quad \] \[ \int_{\mathbb{R}} f_n {\text d}\mu \le \int_{\mathbb{R}} y_n \chi_{E_n} {\text d}\mu = y_n \int_{\mathbb{R}} \chi_{E_n} {\text d}\mu = y_n \mu \left( E_n \right) = 0 . \] From the Monotone Convergence Theorem, \[ \int_{\mathbb{R}} |f|\,{\text d}\mu = \int_{\mathbb{R}} \left( \lim_{n\to\infty} \right) {\text d}\mu = \lim_{n\to\infty} \int_{\mathbb{R}} f_n {\text d}\mu = \lim_{n\to\infty} 0 = 0 . \]

   
Example 2: Let us define function that is zero almost everywhere \[ f(x) = \begin{cases} 1 & \text{if } x = 0 \\ 0 & \text{otherwise.} \end{cases} \] This function is measurable; it is nonzero only at one point, which is a set of measure zero. So f(x) = 0 almost everywhere. The Lebesgue integral is: \[ \int_{\mathbb{R}} |f(x)| \, dx = \int_{\mathbb{R}} \chi_{\{0\}}(x) \, {\text d}x = 0 . \] This confirms the forward direction: if f(x) = 0 almost everywhere, then ∫ |f| = 0.

Let’s define: \[ f(x) = \begin{cases} x & \text{if } x \in \mathbb{Q} \\ 0 & \text{otherwise.} \end{cases} \] This function is measurable because both ℚ and its complement are measurable. The set where f(x) ≠ 0 is countable (the rationals), hence has measure zero. So f(x) = 0 almost everywhere. Its Lebesgue integral is: \[ \int_{\mathbb{R}} |f(x)| \, dx = \sum_{q \in \mathbb{Q}} |q| \cdot \mu(\{q\}) = \sum_{q \in \mathbb{Q}} |q| \cdot 0 = 0 \] Again, this confirms the reverse direction: if ∫ |f| = 0, then f = 0 almost everywhere.

Let us consider function not zero almost everywhere \[ f(x) = \begin{cases} 1 & \text{if } x \in [0,1] \\ 0 & \text{otherwise} \end{cases} \] This function is measurable and nonzero on a set of positive measure. Its integral is: \[ \int_{\mathbb{R}} |f(x)| \, dx = \int_0^1 1 \, dx = 1 \] So ∫ |f| ≠ 0, and indeed f(x) ≠ 0 on a set of positive measure. This confirms the contrapositive of Theorem 2.    ■

End of Example 2

Theorem 2 claims: If a measurable function f : ℝ ↦ [-∞, ∞] has zero integral of its absolute value, then it must be zero almost everywhere — and vice versa. It is one way of saying that integration ‘ignores’ what happens to the integrand on any chosen set of measure 0. Here is a result that says that in way that is often used.

Theorem 3: Let f : ℝ ↦ [−∞, ∞] be an integrable function and g : ℝ ↦ [−∞, ∞] be a Lebesgue measurable function with f(x) = g(x) almost everywhere. Then g must also be integrable and \( \displaystyle \quad \int_{\mathbb{R}} f\,{\text d}\mu = \int_{\mathbb{R}} g\,{\text d}\mu . \)
Let E = { x ∈ ℝ : f(x) ≠ g(x) } and note that f(x) = g(x) almost everywhere means μ(E) = 0.

Write f = (1 − χE)f + χEf. Note that both (1 − χE)f and χEf are integrable because they are measurable and satisfy    |(1 − χE)f| ≤ |f| and |χEf| ≤ |f|. Also \[ \left\vert \int_{\mathbb{R}} \chi_E f\,{\text d}\mu \right\vert \le \int_{\mathbb{R}} \left\vert \chi_E f \right\vert {\text d}\mu = 0 \] as χEf = 0 almost everywhere. Similarly \( \displaystyle \quad \int_{\mathbb{R}} \chi_E g \,{\text d}\mu = 0 . \)

So \begin{align*} \int_{\mathbb{R}} f\,{\text d}\mu &= \int_{\mathbb{R}} \left( \left( 1 - \chi_{E} \right) f + \chi_E f \right) {\text d}\mu \\ &= \int_{\mathbb{R}} \left( 1 - \chi_{E} \right) f \,{\text d}\mu + \int_{\mathbb{R}} \chi_E f \, {\text d}\mu \\ &= \int_{\mathbb{R}} \left( 1 - \chi_{E} \right) f \,{\text d}\mu + 0 \\ &= \int_{\mathbb{R}} \left( 1 - \chi_{E} \right) g \,{\text d}\mu . \end{align*} The same calculation (with |f| in place of f) shows \[ \int_{\mathbb{R}} \left( 1 - \chi_{E} \right) |g| \,{\text d}\mu = \int_{\mathbb{R}} |f|\,{\text d}\mu < \infty , \] so that (1 − χE)g must be integrable. Thus, g = (1 − χE) g + χEg s also integrable (because \( \displaystyle \quad \int_{\mathbb{R}} \left\vert \chi_E \,g \right\vert {\text d}\mu = 0 \quad \) and so χEg is integrable and g is then the sum of two integrable functions). Hence, we have \[ \int_{\mathbb{R}} g\,{\text d}\mu = \int_{\mathbb{R}} \left( 1 - \chi_E \right) g\,{\text d}\mu + \int_{\mathbb{R}} \chi_E g\,{\text d}\mu = \int_{\mathbb{R}} f\,{\text d}\mu + 0 . \]

   
Example 3: Let’s walk through a few illustrative examples to bring this theorem to life. First, we show that changing a function on a measure-zero set does not affect its integrability. Let’s define the function \[ f(x) = e^{-x^2} \qquad \text{for all } x \in \mathbb{R} . \] Now define another function \[ g(x) = \begin{cases} \infty & \text{if } x = 0 , \\ e^{-x^2} & \text{otherwise.} \end{cases} \] Function f is clearly integrable over ℝ with \( \displaystyle \quad \int_{\mathbb{R}} f(x) \, {\text d}x = \sqrt{\pi} . \quad \) Function g differs from f only at one point, x = 0, which is a measure-zero set. So f(x) = g(x) almost everywhere. Hence, by Theorem 3: g is also integrable and \( \displaystyle \quad \int_{\mathbb{R}} g(x) \, dx = \int_{\mathbb{R}} f(x) \, dx = \sqrt{\pi} . \)

Now we consider the case when one function is differing on a countable set compared to another function. Let f(x) = 0 for all x ∈ ℝ, and define: \[ g(x) = \begin{cases} n & \text{if } x = q_n \text{ (the } n\text{th rational in } \mathbb{R}) \\ 0 & \text{otherwise} . \end{cases} \] Hence, f is integrable with ∫ℝ f(x) dx = 0. Function g is nonzero only on the countable set of rationals, which has measure zero So f(x) = g(x) almost everywhere. Therefore, g is integrable and ∫ℝ g(x) dx = 0.

Now we consider the case when functions not equal almost everywhere. Let \[ f(x) = 0, \quad g(x) = 1 \text{ on } [0,1] \] Function f is integrable with ∫ f = 0. Function g is integrable with ∫ g = 1. But ff(x) ≠ g(x) on a set of positive measure (namely, all of [0,1]) So Theorem 3 does not apply, and indeed ∫ f ≠ ∫ g.    ■

End of Example 3
Beppo Levi’s monotone convergence theorem (MCT): Let (X, 𝔗, µ) be a measure space, for instance, ℝ or ℝd with the Borel σ-algebra and the Lebesgue measure. Let ( fₙ )n∈ℕ be a sequence of non-negative measurable functions on X. Assume that ( fₙ )n∈ℕ are increasing for all x ∈ X. Then the function f : X → ℝ ∪ {+∞}, defined by \[ f(x) = \lim_{n\to\infty} f_n (x) \qquad\mbox{for all}\quad x \in X, \] is measurable and we have \[ \lim_{n\to\infty} \int_X f_n {\text d}\mu = \int_X f\,{\text d}\mu \] (this quantity may be either finite or infinite).
Since fₙ ↗ f, then fₙ &le f almost everywhere on E. It follows from the monotonicity of the integral of nonnegative functions that \( \displaystyle \quad \int_E f_n {\text d}x \le \int_E f\, {\text d}x \quad \) and so \( \displaystyle \quad \limsup_{n\to\infty} \int_E f_n {\text d}x \le \int_E f\, {\text d}x .\quad \) On the other hand, it follows from Fatou’s Lemma that \( \displaystyle \quad \int_E f\,{\text d}x \le \liminf_{n\to\infty} \int_E f_n {\text d}x . \quad \) Hence, \( \displaystyle \quad \int_E f\,{\text d}x = \lim_{n\to\infty} \int_E f_n {\text d}x . \)
   
Example 4: Let us define a sequence of functions fₙ : [0,1] → ℝ by \[ f_n(x) = \min(n, 1) \] So for x ∈ [0,1], fₙ(x) = 1 for all n ≥ 1. The sequence is constant, hence trivially increasing. The pointwise limit is f(x) = 1

Then \[ \int_0^1 f_n(x) \, {\text d}x = \int_0^1 1 \, {\text d}x = 1 \quad \text{for all } n, \] and limits of integrals become \[ \lim_{n \to \infty} \int_0^1 f_n(x) \, {\text d}x = 1 = \int_0^1 \lim_{n \to \infty} f_n(x) \, {\text d}x . \] Therefore, the MCT holds: the limit of the integrals equals the integral of the limit.

Example 2: Infinite limit.

Let us define fₙ(x) = χ{[0,n]}(x) on ℝ, the characteristic function of the interval [0,n]. Then:

Each fₙ is measurable and non-negative. fₙ(x) ≤ fn+1(x) for all x. The pointwise limit is \[ f(x) = \lim_{n \to \infty} f_n(x) = \chi_{[0,\infty)}(x) . \] Now compute: \[ \int_{\mathbb{R}} f_n(x) \, {\text d}x = \int_0^n 1 \, {\text d}x = n , \] and \begin{align*} \lim_{n \to \infty} \int f_n(x) \, {\text d}x &= \infty , \\ \int_{\mathbb{R}} f(x) \, {\text d}x &= \int_0^\infty 1 \, {\text d}x = \infty . \end{align*} Again, MCT holds: both sides equal \infty.

Finally, we consider the example where the sequence is not monotone. Let fₙ(x) = χ[0,1/n](x), the characteristic function of the interval [0,1/n]. Then

fₙ(x) → 0 pointwise. But fₙ(x) is not increasing — it’s decreasing So MCT does not apply. In this case: \[ \int f_n(x) \, {\text d}x = \frac{1}{n} \to 0, \quad \int \lim f_n(x) \, {\text d}x = \int 0 \, {\text d}x = 0 . \] Although the conclusion still holds here, it’s not guaranteed by MCT — instead, this is a case for the Dominated Convergence Theorem.    ■
End of Example 4

The Monotone Convergence Theorem is primarily associated with the French mathematician Henri Lebesgue (1875--1941) and the Italian mathematician Beppo Levi (1875--1961). Lebesgue first established the theorem for functions, and Levi later provided a generalization to a more general measure-theoretic setting. This result is commonly used both to prove the integrability of a function obtained by a limit process, and as a way to compute the integral of such functions.

Fatou's Lemma: Let fn : ℝ → [0, ∞] be (nonnegative) Lebesgue measurable functions. Then \[ \int_{\mathbb{R}} \liminf_{n\to \infty} f_n {\text d} \mu \leqslant \liminf_{n\to \infty} \int_{\mathbb{R}} f_n {\text d} \mu . \]
Let gₙ(x) = infk≥nfk(x), so we have \[ \left( \liminf_{n\to\infty} f_n \right) (x) = \liminf_{n\to\infty} f_n (x) = \lim_{n\to\infty} \left( \int_{k\ge n} f_k (x) \right) = \lim_{n\to\infty} g_n (x) . \] Note that gₙ(x) = infk≥nfk(x) ≤ infk≥n+1fk(x) = gn+1(x) so the sequence { gn(x) ]n≥1 is monotone increasing for each x and so the monotone convergence theorem says that \[ \lim_{n\to\infty} \int_{\mathbb{R}} g_n {\text d}x = \int_{\mathbb{R}} \lim_{n\to\infty} g_n {\text d}x = \int_{\mathbb{R}} \liminf_{n\to\infty} f_n {\text d}\mu . \] However, also gₙ &le' fk for each k ≥ n and so \[ \int_{\mathbb{R}} g_n \,{\text d}\mu \le \int_{\mathbb{R}}f_k {\text d}\mu \qquad (k\ge n) \] or \[ \int_{\mathbb{R}} g_n \,{\text d}\mu \le \inf_{k\ge n} \int_{\mathbb{R}}f_k {\text d}\mu . \] Hence, \[ \liminf_{n\to\infty} \int_{\mathbb{R}} f_n \,{\text d}\mu = \lim_{n\to\infty} \left( \inf_{k\ge n} \int_{\mathbb{R}} f_k \,{\text d}\mu \right) \ge \lim_{n\to\infty} \int_{\mathbb{R}} g_n \,{\text d}\mu = \int_{\mathbb{R}} \liminf_{n\to\infty} f_n {\text d}\mu . \]
   
Example 5: Let us take the characteristic function, fₙ = χ[n,2n], and notice that \( \displaystyle \quad \int_{\mathbb{R}} f_n {\text d}\mu = n \to \infty \quad \) as n → ∞, but for each x ∈ ℝ, \( \displaystyle \quad \lim_{n\to\infty} f_n (x) = 0 . \quad \) So \[ \liminf_{n\to\infty} \int_{\mathbb{R}} f_n {\text d}\mu = \infty \ > \ \int_{\mathbb{R}} \liminf_{n\to\infty} f_n {\text d}\mu = \int_{\mathbb{R}} 0\, {\text d}\mu = 0 . \] This also shows that the Monotone Convergence Theorem is not true without ‘Monotone’.    ■
End of Example 5
    Even when dealing with “concrete” functions, Lebesgue’s theory is extremely useful because of the power of its general statements about exchanging limits and integrals. The main theorem is Lebesgue’s dominated convergence theorem.
Lebesgue’s dominated convergence theorem (DCT): Let (X, 𝔗, µ) be a measure space, for instance, ℝ or ℝd with the Borel σ-algebra and the Lebesgue measure. Let ( fₙ )n∈ℕ be a sequence of complex-valued measurable functions on X. Assume that ( fₙ )n∈ℕ converges pointwise almost everywhere to a function f : X → ℂ, and moreover assume that there exists a µ-integrable function g : X → ℝ+ such that |fₙ| ≤ g almost everywhere on X, for all n ∈ ℕ. Then
  1. f is µ-integrable;
  2. for any measurable subset A ⊂ X , we have \[ \lim_{n\to\infty} \int_A f_n {\text d}\mu = \int_A \lim_{n\to\infty} \,f_n {\text d}\mu = \int_A g \,{\text d}\mu \]
Since |fₙ(x)| ≤ g(x) and g is integrable, \( \quad \int_{\mathbb{R}} \left\vert f_n \right\vert {\text d}\mu \leq \int_{\mathbb{R}} g\,{\text d}\mu < \infty . \quad \) Therefore, fₙ is integrable for each n. We know that f is measurable (as a pointwise limit of measurable functions) and then, similarly, \[ | f(x) | = \lim_{n\to\infty} \left\vert f_n (x) \right\vert \le g(x) \] implies that f is integrable too.

The proof does not work properly if g(x) = ∞ for some x. We know that g(x) < ∞ almost everywhere. So we can take E = {x ∈ ℝ : g(x) = ∞} and multiply g and each of the functions fₙ and f by 1 − χE to make sure all the functions have finite values. As we are changing them all only on the set E of measure 0, this change does not affect the integrals or the conclusions. We assume then all functions have finite values.

Let hₙ = g − fₙ, so that hₙ ≥ 0. By Fatou’s lemma \[ \liminf_{n\to\infty} \int_{\mathbb{R}} \left( g - f_n \right) {\text d}\mu \ge \int_{\mathbb{R}} \liminf_{n\to\infty} \left( g - f_n \right) {\text d}\mu = \int_{\mathbb{R}} \left( g - f \right) {\text d}\mu \] and that gives \[ \liminf_{n\to\infty} \left( \int_{\mathbb{R}} g\,{\text d}\mu - \int_{\mathbb{R}} f_n\,{\text d}\mu \right) = \int_{\mathbb{R}} g\,{\text d}\mu - \liminf_{n\to\infty} f_n {\text d}\mu \ge \int_{\mathbb{R}} g\,{\text d}\mu - \int_{\mathbb{R}} f\,{\text d}\mu \] or \[ \liminf_{n\to\infty} \int_{\mathbb{R}} f_n {\text d}\mu \le \int_{\mathbb{R}} f\, {\text d}\mu . \] Repeat this Fatou’s lemma argument with g + fₙ rather than g − fₙ. We get \[ \liminf_{n\to\infty} \int_{\mathbb{R}} \left( g + f_n \right) {\text d}\mu \ge \int_{\mathbb{R}} \liminf_{n\to\infty} \left( g + f_n \right) {\text d}\mu = \int_{\mathbb{R}} \left( g + f \right) {\text d}\mu \] and that gives \begin{align*} \liminf_{n\to\infty} \left( \int_{\mathbb{R}} g\,{\text d}\mu + \int_{\mathbb{R}} f_n\,{\text d}\mu \right) &= \int_{\mathbb{R}} g\,{\text d}\mu + \liminf_{n\to\infty} \int_{\mathbb{R}} f_n\,{\text d}\mu \\ &\ge \int_{\mathbb{R}} g\,{\text d}\mu + \int_{\mathbb{R}} f\,{\text d}\mu \end{align*} or \[ \liminf_{n\to\infty} \int_{\mathbb{R}} f_n\,{\text d}\mu \ge \int_{\mathbb{R}} f\,{\text d}\mu . \] Combining with the previous inequality, we have \[ \int_{\mathbb{R}} f\,{\text d}\mu \le \liminf_{n\to\infty} \int_{\mathbb{R}} f_n\,{\text d}\mu \le \limsup_{n\to\infty} \int_{\mathbb{R}} f_n\,{\text d}\mu \le \int_{\mathbb{R}} f\,{\text d}\mu , \] which forces \[ \int_{\mathbb{R}} f\,{\text d}\mu = \liminf_{n\to\infty} \int_{\mathbb{R}} f_n\,{\text d}\mu = \limsup_{n\to\infty} \int_{\mathbb{R}} f_n\,{\text d}\mu \] and that gives the result because if \( \displaystyle \quad \liminf_{n\to\infty} a_n = \limsup_{n\to\infty} a_n \quad \) (for a sequence (𝑎ₙ), it implies that \( \displaystyle \quad \lim_{n\to\infty} a_n \quad \) exists and \( \displaystyle \quad \lim_{n\to\infty} a_n = \limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n . \)

   
Example 6: Let \( \displaystyle \quad f_n (x) = \frac{n}{1 + n^2 x^2} . \quad \) This is the Cauchy kernel often used in approximation theory.

Pointwise limit exists: \( \displaystyle \quad f_n (x) \to 0 \quad \) for all x ≠ 0. The inetgral for each n is \[ \int_{-\infty}^{\infty} f_n (x) \,{\text d} x = \pi . \]

RJB
However, \[ \int_{-\infty}^{\infty} \lim_{n\to\infty} f_n (x) \,{\text d} x = 0 . \] There is no integrable function g(x) such that |fₙ(x)| ≤ g(x) for all n. In fact, the peak of fₙ at x = 0 becomes increasingly sharp and tall, concentrating mass near zero. There is a classic example of concentration without domination.    ■
End of Example 6
    When dealing with improper integrals, you must analyze the double limit carefully to ensure the convergence is uniform:
\[ \lim_{n\to\infty} \int_{\mathbb{R}} f_n (x)\,{\text d}x = \lim_{n\to\infty} \lim_{R\to +\infty} \int_{-R}^R f_n (x)\,{\text d}x = \lim_{R\to +\infty} \lim_{n\to\infty} \int_{-R}^R f_n (x)\,{\text d}x . \]
When the dominated convergence is uniform, |fₙ(x)| ≤ g(x) for all n amd almost every x, it is safe to exchange the limits. The dominating function must be integrable over the entire domain. If the integral is improper, you are really applying the DCT to a truncated domains and then taking a limit---so interchanging limits must be justified.    
Example 7: Let \( \displaystyle \quad f_n (x) = \frac{x^2}{1 + n\,x^4} . \quad \) For each fixed x, fₙ(x) → 0 as n → ∞. Observe that \[ \left\vert f_n (x) \right\vert \leq \frac{x^2}{ + x^4} =: g(x) . \] The dominating function g(x) is integrable over whole ℝ. So we make a conclusion that the DCT is applicable and \[ \lim_{n\to\infty} \int_{-\infty}^{\infty} f_n (x)\,{\text d}x = \lim_{n\to\infty} \lim_{R, S\to \infty} \int_{-S}^R f_n (x) \,{\text d}x = \lim_{R, S\to +\infty} \lim_{n\to\infty} \lim_{n\to\infty} \int_{-S}^R f_n (x) \,{\text d}x = 0 . \] Here the double limit \[ \lim_{n\to\infty} \lim_{R\to \infty} \int_{-R}^R f_n (x) \,{\text d}x \] is justified because the integrals are uniformly dominated by an integrable function.    ■
End of Example 7
    The dominated convergence theorem works on unbounded domains if you can find a fixed integrable function that dominates all fₙ(x). If the sequence concentrates mass (e.g., like a delta sequence), domination may fail, and DCT is not applicable. In such cases, interchanging limits is invalid, and you must analyze the double limit carefully.

Here finally is the version of the dominated convergence theorem adapted to a series of functions:

(Term by term integration of a series) Theorem 5: Let (X, 𝔗, µ) be a measure space, for instance, ℝ or ℝd with the Borel σ-algebra and the Lebesgue measure. Let \( \displaystyle \quad \sum f_n \quad \) be a series of complex-valued measurable functions on X such that \[ \sum_{n=0}^\infty \int | f_n | \,{\text d}\mu < +\infty . \] Then the series of functions \( \displaystyle \quad \sum f_n \quad \) converges absolutely almost everywhere on X and its sum is μ-integrable. Moreover, its integral may be computed by term by term integration: we have \[ \int \sum_{n=0}^\infty f_n {\text d}\mu = \sum_{n=0}^\infty \int f_n {\text d}\mu . \]
   
Example 8: The above theorem is a version of the term-by-term integration theorem for series of measurable functions, often called the Fubini–Tonelli-type result for series. It guarantees that if the sum of the integrals of the absolute values of the terms is finite, then:
  • The series ∑n≥0 fₙ(x) converges absolutely almost everywhere.
  • The sum function f(x) = ∑n≥0 fₙ(x) is integrable.
  • The integral of the sum equals the sum of the integrals: \[ \int_X \left( \sum_{n=0}^\infty f_n(x) \right) {\text d}\mu = \sum_{n=0}^\infty \int_X f_n(x) \, {\text d}\mu . \]
Let’s look at two illustrative examples.

Example 1: Geometric series of functions on [0,1]: \[ f_n(x) = \frac{x^n}{2^n} \qquad \mbox{on } [0,1], \] with Lebesgue measure μ.

  • Each fₙ is measurable and non-negative.
  • For each x ∈ [0,1], the series \( \displaystyle \quad \sum_{n=0}^\infty f_n(x) = \sum_{n=0}^\infty \frac{x^n}{2^n} \quad \) converges absolutely.
  • The pointwise sum is: \[ f(x) = \sum_{n=0}^\infty \frac{x^n}{2^n} = \frac{1}{1 - \frac{x}{2}} = \frac{2}{2 - x}, \quad \text{for } x \in [0,1) \]
Now compute the integral of each term: \[ \int_0^1 f_n(x) \, {\text d}x = \int_0^1 \frac{x^n}{2^n} {\text d}x = \frac{1}{2^n(n+1)} . \] So the sum of the integrals is: \[ \sum_{n=0}^\infty \int_0^1 f_n(x) \, dx = \sum_{n=0}^\infty \frac{1}{2^n(n+1)} < \infty . \] All conditions of Theorem 7 are satisfied. Therefore, \[ f(x) = \sum f_n(x) is integrable on [0,1] \] and \[ \int_0^1 f(x) \, {\text d}x = \sum_{n=0}^\infty \int_0^1 f_n(x) \, {\text d}x . \]

Example 2: Alternating series with complex-valued functions.

Let \( \displaystyle \quad f_n(x) = \frac{(-1)^n}{n} \chi_{[0,1]}(x), \quad \) where χ[0,1] is the indicator function of [0,1].

  • Each fₙ is measurable and supported on [0,1].
  • The pointwise series is \[ f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n} \chi_{[0,1]}(x) = \begin{cases} -\ln(2) & \text{if } x \in [0,1] \\ 0 & \text{otherwise} \ln(2) & \text{if } x \in [0,1] . \end{cases} The absolute integrals: \[ \sum_{n=1}^\infty \int_{\mathbb{R}} |f_n(x)| \, {\text d}x = \sum_{n=1}^\infty \frac{1}{n} \cdot \mu([0,1]) = \sum_{n=1}^\infty \frac{1}{n} = \infty . \] So this does not satisfy the hypothesis of Theorem 7 — the sum of the integrals of |fₙ| diverges. Hence, term-by-term integration is not justified here.

    Example 3: Decaying sine series.

    Let \( \displaystyle \quad f_n(x) = \frac{\sin(n x)}{n^2} on [0, \pi] . \quad \) Each fₙ is measurable and bounded. The series ∑ fₙ(x) converges absolutely for all x, since \( \displaystyle \quad \sum \frac{1}{n^2} < \infty . \quad \) The integral of each term: \[ \int_0^\pi \frac{\sin(n x)}{n^2} {\text d}x = \frac{1 - (-1)^n}{n^2} . \] So \[ \sum_{n=1}^\infty \left| \int_0^\pi \frac{\sin(n x)}{n^2} dx \right| \leq \sum_{n=1}^\infty \frac{2}{n^2} < \infty \] The series of integrals of |fₙ| converges, so Theorem 7 applies: \[ \int_0^\pi \sum_{n=1}^\infty \frac{\sin(n x)}{n^2} dx = \sum_{n=1}^\infty \int_0^\pi \frac{\sin(n x)}{n^2} \,{\text d}x . \]    ■

    End of Example 8

 

 

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