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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 { fn(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 fn(x) eluated 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 { fn(x) }.

A sequence { fn(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 { fn } does not converge pointwise to f has zero measure, i.e., μ(E) = 0.
Similarly, we say that the sequence { fn } 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:    ■
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:    ■
End of Example 2

Theorem 2 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

   
Example 2:    ■
End of Example 3
Monotone convergence theorem:
   
Example 5:    ■
End of Example 5
Fatou's Lemma: Let fn : ℝ → [0, ∞] be (nonnegative) Lebesgue measurable functions. Then \[ \int_{\mathbb{R}} \liminf_{n\to \infty} f_n {\text d} \mu \le \liminf_{n\to \infty} \int_{\mathbb{R}} f_n {\text d} \mu . \]
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 eqch x and so the monotone convergence theorem says that \[ \lim_{n\to\infty} \int_{\mathbb{R}} g_n {\text d}x = \int_{\mathbb{R}} \lim_{n\to\infty} g_n {\text d}x = \int_{\mathbb{R}} \liminf_{n\to\infty} f_n {\text d}\mu . \]
   
Example 5:    ■
End of Example 5
Monotone convergence theorem:
   
Example 5:    ■
End of Example 5

 

 

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