Example 1: Let us consider the function \[ f(x) = \frac{1}{x}\,\sin \left( \frac{1}{x} \right) \quad \mbox{on}\quad (0, \pi ) . \] This function is not absolutely integrable because near the origin f(x) ∼ 1/x. So \[ \int_0^{\pi} \left\vert f(x) \right\vert {\text d} x = \infty . \] Mathematica provides a huge numerical value that indicates its divergence:

But f(x) is integrable, as Mathematica confirms:

• x = 1/t,
• dt = −dt/t²,
• as x → 0+0 ⇔ t ↓ 0, t → +∞.

This example shows that the oscillation near the singular point is strong enough to overcome the 1/x blow-up.

This example also show that the given function f(x) = sin(1/x)/x is not in 𝔏¹. Therefore, the classical (based on improper Riemann integration) Fourier theory is wider than modern theory based on Lebesgue integration.

We can expand this function into sine-Fourier series \[ \frac{1}{x}\,\sin \left( \frac{1}{x} \right) = \sum_{n\ge 1} b_n \sin (nx) . \] With Mathematica, we can evaluate several coefficients, numerically. The following figure presents graph of function f(x) in blue, and its 20-term sine0Fourier approximation, in red.

Example 2: Let us define on (−π, π) the function \[ f(x) = \begin{cases} \frac{\sin (1/x)}{x\,\ln (1/|x|)} , &\quad 0 < |x| < e^{-2} , \\ 0, &\quad e^{-2} \le |x| \le \pi . \end{cases} \] This function is extended in odd way so its Fourier series will contain only sine functions. As usual, f is assumed to be 2π periodic because we extend it into the Fourier series.

The given function is not absolutely integrable because near zero, \[ |f(x)| \,\sim \, \frac{1}{|x|\,\ln (1/|x|)} . \] So \[ \int_0^{1/e^2} |f(x)|\,{\text d}x \ge \int_0^{1/e^2} \frac{{\text d}x}{|x|\,\ln (1/|x|)} . \] Substitute t = ln(1/x), then dx = −e^−tdt, and we get \[ \int_0^{1/e^2} |f(x)|\,{\text d}x \ge \int_2^{+\infty} \frac{{\text d}t}{t} = \infty . \] So f ∉ 𝔏¹([−π, π). However, the Fourier sine coefficients exist: \[ b_n = \frac{2}{\pi} \int_0^{\pi} f(x)\,\sin (nx)\,{\text d}x = \frac{2}{\pi} \int_0^{1/e^2} \,\frac{\sin (1/x)}{x\,\ln (1/x)}\,\sin (nx)\,{\text d}x . \] Let t = 1/x, then dx = −dt/t² and ln(1/x) = ln(t). Then \[ b_n = \frac{2}{\pi} \int_{e^2}^{\infty} \frac{\sin t}{\left( 1/t \right) \ln t}\,\sin \left( \frac{n}{t} \right) \frac{{\text d}t}{t^2} = \frac{2}{\pi} \int_{e^2}^{\infty} \frac{\sin t\,\sin (n/t)}{t\,\ln t} \,{\text d}t . \] For large t, \[ \sin \left( \frac{n}{t} \right) \,\sim \,\frac{n}{t} , \] so the integrand behaves like \[ \frac{\sin t}{t\,\ln t} \cdot \frac{n}{t} = n\,\frac{\sin t}{t^2 \ln t} . \] Now \[ \int_{e^2}^{\infty} \left\vert \frac{\sin t}{t^2 \ln t} \right\vert {\text d}t \le \int_{e^2}^{\infty} \frac{{\text d}t}{t^2 \ln t} < \infty . \] Hence, the integral defining bₙ converges absolutely for each n. Thus, all Fourier coefficients exist, even so f ∉ 𝔏¹.

However, the corresponding Fourier series converges very slow because bₙ ∼ 1/lnn (see section).    ■

Example 3: Thomae's function is a real-valued function of a real variable that is named after Carl Johannes Thomae, but has many other names: the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function (not to be confused with the integer ruler function), the Riemann function, or the Stars over Babylon (John Horton Conway's name). It can be defined as: \[ f(x) = \begin{cases} 1/q , \qquad & x = \frac{p}{q} \quad\mbox{in lowest terms}, \\ 0, \qquad & x \notin \mathbb{Q} . \end{cases} \] This function has a remarkably rich structure

• Thomae's function 𝑓 is bounded and maps all real numbers to the unit interval, 𝑓 : ℝ → ]0,1].
• 𝑓 is periodic with period 1 : 𝑓 ( x + n ) = 𝑓 ( x ) for all integers n and all real x.
• 𝑓 is discontinuous at every rational number, so its points of discontinuity are dense within the real numbers.
• 𝑓 is continuous at every irrational number, so its points of continuity are dense within the real numbers.
• 𝑓 is nowhere differentiable.
• 𝑓 has a proper local maximum at each rational number, providing an example of a function with a dense set of proper local maxima.
• 𝑓 is Riemann integrable on any interval and the integral evaluates to 0 over any set.

Now we determine the upper Riemann sum. For any given ε > 0, there are only finitely many rational numbers in whose denominator 1/q *ge; ε/2. Let k be the number of such points.

We can choose a partition P where each of these k points is enclosed in a very narrow subinterval of total combined length less than &eosilon;/2. On these narrow subintervals, the maximum value of 𝑓(𝑥) is bounded above by 1. On all other subintervals, the maximum value of 𝑓(𝑥) is strictly less than ε/2.

The upper Riemann sum U(𝑓, P) can be split into two parts: \[ U(f, P) < \left( 1 \cdot \frac{\varepsilon}{2} \right) + \left( \frac{\varepsilon}{2} \cdot 1 \right) = \varepsilon . \] Since ε can be chosen to be arbitrarily small, the infimum of the upper Riemann sums (the upper Riemann integral) must be 0. So the integral must be zero.

Alternatively, from the perspective of Lebesgue integration, Thomae's function is equal to almost everywhere because the set of rational numbers has a Lebesgue measure of zero. The Lebesgue integral of any function that vanishes almost everywhere is 0.

So all Fourier coefficients of Thomae's function are zeroes. Therefore, the Fourier analysis is not suitable for Thomae's function. :    ■

Example 10: We consider the power function \( f(x) = \left( x^2 - \ell^2 \right)^2 . \) Expanding it into Fourier series, we get

Example 10B: Here is another qadratic function (on interval −ℓ < x < ℓ):

Example 4:

For example, let us consider the unbounded function 𝑓(𝑥) = 1/√𝑥 on interval [0, π]. For this unbounded function, the integral at the singularity 𝑥 = 0 is evaluated as an improper Riemann integral. The Riemann-Lebesgue Lemma still applies perfectly because the function is absolutely integrable over the interval: \[ \int_0^{\pi} \, \left\vert \frac{1}{\sqrt{x}} \right\vert {\text d} x = 2 \sqrt{\pi} < \infty . \] When expanding a function given only on the interval [0, π], it is standard practice to extend it to the interval [−π, π] to form either a Fourier Cosine Series (even extension) or a Fourier Sine Series (odd extension).

Option A: The Fourier Cosine Series The even extension creates a series composed entirely of cosine terms: \[ f(x) = \frac{1}{2}\, a_0 + \sum_{n\ge 1} \, a_n\,\cos (nx) . \] The DC coeficient: \[ a_0 = \frac{2}{\pi}\,\int_0^{\pi}\, \frac{1}{\sqrt{x}} \, {\text d} = \frac{2}{\pi}\,\left[ 2\sqrt{x} \right]_{x=0}^{x= \pi} = \frac{4}{\sqrt{\pi}} . \] The AC coefficients: \[ a_n = \frac{2}{\pi}\,\int_0^{\pi}\, \frac{1}{\sqrt{x}} \,\cos (nx)\, {\text d} = . \] Option B: The Fourier Sine Series The odd extension creates a series composed entirely of sine terms: \[ f(x) = \sum_{n\ge 1}\, b_n \,\sin (nx) , \] where \[ b_n = \frac{2}{\pi}\,\int_0^{\pi}\, \frac{1}{\sqrt{x}} \,\sin (nx)\, {\text d} = . \] All these integrals are expressed through Fresnel Integrals \[ C(z) = \int_0^z \,\cos \left( \frac{\pi}{2}\,r^2 \right) {\text d}r , \qquad S(z) = \int_0^z \,\sin \left( \frac{\pi}{2}\,r^2 \right) {\text d}r . , \] Evaluating these integrals requires a change of variables \( \displaystyle \quad u = \sqrt{\frac{2nx}{\pi}} , \quad \) which maps then directly to the classical Fresnel Integrals. Solving the integrals yields the exact formulas for the coefficients: \begin{align*} a_n &= \frac{2\sqrt{2}}{\sqrt{\pi n}}\, C \left( \sqrt{2n} \right) , \\ b_n &= \frac{2\sqrt{2}}{\sqrt{\pi n}}\, S \left( \sqrt{2n} \right) . \end{align*} To see how these coefficients behave as the frequency n grows, we take the limit using the known asymptotic behavior of the Fresnel integrals. As their arguments roll out to infinity, both Fresnel functions stabilize and converge to exactly fffffffrac12;: \[ \lim_{n\to\infty} \ C \left( \sqrt{2n} \right) = \frac{1}{2} \qquad \mbox{and}\qquad \lim_{n\to\infty} \ S \left( \sqrt{2n} \right) = \frac{1}{2} \] Substituting these asymptotic limits into our coefficient equations reveals their large-n behavior: \[ a_n = \frac{2\sqrt{2}}{\sqrt{\pi n}} \left( \frac{1}{2} \right) = \frac{\sqrt{2}}{\sqrt{\pi n}} , \qquad b_n = \frac{2\sqrt{2}}{\sqrt{\pi n}} \left( \frac{1}{2} \right) = \frac{\sqrt{2}}{\sqrt{\pi n}} . \] Obviously, these coefficients vanish at infinity. Smooth bounded functions have Fourier coefficients that decay incredibly fast. The presence of the vertical asymptote singularity at x = 0 drops the decay rate down to 1/√n.

To explain why the lemma remains valid despite the vertical asymptote, you can use a simple truncation technique that relies only on basic limits:

1. Isolate the Singularity: Split the integral into a tiny region near the singularity [0, ε], and a safe, well-behaved region [ε, π]: \[ \int_0^{\pi} \,\frac{\sin (nx)x}{\sqrt{x}} \,{\text d}x = \int_0^{\varepsilon} \,\frac{\sin (nx)x}{\sqrt{x}} \,{\text d}x + \int_{\varepsilon}^{\pi} \,\frac{\sin (nx)x}{\sqrt{x}} \,{\text d}x . \]
2. Bound the Problematic Area: On the tiny interval , use the fact that |sin(nx)| to prove that the area is bounded entirely by the choice ofε regardless of how large n becomes: \[ \left\vert \int_0^{\varepsilon} \, \frac{\sin (nx)}{\sqrt{x}}\,{\text d}x \right\vert \le \int_0^{\varepsilon} \, \frac{1}{\sqrt{x}} \,{\text d}x = 2\sqrt{\varepsilon} . \]
3. Apply the Bounded Lemma to the Rest: On the remaining interval [ε, π], the function is bounded and properly Riemann integrable. Therefore, the classical Riemann-Lebesgue Lemma holds exactly: \[ \lim_{n\to\infty}\ \int_{\varepsilon}^{\pi} \,\frac{\sin (nx)x}{\sqrt{x}} \,{\text d}x = 0 . \]
4. Combine the Limits: By making ε arbitrarily small, the first part shrinks to zero. By letting n → ∞, the second part drops to zero. Consequently, the total integral is forced to zero    ■
   End of Example 4

Example 5: We present a few examples where direct application of the Riemann-Lebesgue lemma is not possible.

Example 5A: Consider \[ f(x) = \frac{\cos (1/x^2 )}{\sqrt{x}} , \qquad 0 < x \le 1 . \] First, we show that 𝑓 is improperly Riemann integrable. Indeed, \( \displaystyle \quad \int_0^1 \,\frac{\cos (1/x^2 )}{\sqrt{x}} \,{\text d}x \quad \) converges conditionally. Setting u = 1/x², we obtain \[ {\text d}x = - \frac{1}{2}\, u^{-3/2} \ {\text d}u, \qquad x^{-1/2} = u^{1/4} , \] so \[ \int_0^1 \,\frac{\cos (1/x^2 )}{\sqrt{x}} \,{\text d}x = \frac{1}{2}\,\int_1^{\infty} \,\frac{\cos u}{u^{5/4}}\,{\text d}u . \] which actually converges absolutely, because \( \displaystyle \quad \int_1^{\infty} \,u^{-5/4}\ {\text d}u < \infty . \quad \) Therefore, 𝑓 ∈ 𝔏¹(0,1). Consequently, the ordinary Riemann–Lebesgue lemma applies.    ▣

Example 5B:

We consider another function \[ g(x) = \frac{\cos (1/x^2 )}{x} , \qquad 0 < x \le 1 . \] This integral converges conditionally by Dirichlet test, \[ \int_0^1 \,\frac{\cos (1/x^2 )}{x} \,{\text d}x = \frac{1}{2}\, \int_1^{\infty} \,\frac{\cos u}{u}\,{\text d}u < \infty , \] but not absolutely, \[ \frac{1}{2}\, \int_1^{\infty} \,\frac{|\cos u|}{u}\,{\text d}u = \infty . \] Thus,

• 𝑔 is improperly Riemann integrable,
• 𝑔 ∉ 𝔏¹(0,1).

Let \[ c_n = \int_0^1 \,\frac{\cos (1/x^2 )}{x} \ e^{\mathbf{j}nx} \ {\text d}x . \] We will show that cₙ → 0 as n → ∞.

Step 1. Change variables u = 1/x². Then \[ x = u^{-1/2}, \qquad {\text d}x = - \frac{1}{2}\,u^{-3/2}\ {\text d}u , \] and \[ \frac{{\text d}x}{x} = - \frac{1}{2} \,\frac{{\text d}u}{u} . \] Hence, \[ c_n = \frac{1}{2}\,\int_1^{\infty} \,\frac{\cos u}{u}\, e^{-\mathbf{j}\,n/\sqrt{u}}\ {\text d}u . \] Step 2. Split the oscillation using \[ \cos u = \frac{1}{2}\, e^{\mathbf{j}\,u} + \frac{1}{2}\, e^{-\mathbf{j}\,u} . \] It suffices to study \[ I_n^{\pm} = \int_1^{\infty} \, \frac{1}{u}\, e^{\pm\mathbf{j}\,u - \mathbf{j}\,n/\sqrt{u}} \ {\text d}u . \] Step 3. Integration by parts. Consider \[ \phi_n (u) = u - \frac{n}{\sqrt{u}} . \] Then \[ \phi'_n (u) = 1 + \frac{n}{2\,u^{3/2}} \ge 1 . \] Thus the phase has no stationary points.

• the boundary term at u = ∞ is 0,
• at u = 1, \[ \left\vert \frac{1}{\phi'_b (1)} \right\vert = \frac{1}{1 + n/2} \,\to\, 0; \]
• moreover, \[ \frac{\text d}{{\text d}u} \left( \frac{1}{u\,\phi'_n (u)} \right) = O \left( \frac{1}{u^2} \right) , \] uniformly in n, because \[ \phi'_n (u) \ge 1 , \qquad \phi''_n (u) = - \frac{3n}{4\,u^{5/4}} \] Since u⁻² ∈ Lfr'²(1,∞), the remaining integral is bounded by an absolutely convergent integral, and another integration-by-parts estimate (or the uniform version of Dirichlet's test) yields \[ I_n^{+} \,\to\,0 . \]
,/ul> Exactly the same argument applies to \( \displaystyle \quad I_n^{-} \quad \) because \[ \phi_n (u) = -u - \frac{n}{\sqrt{u}} , \qquad \phi'_n (u) = -1 + \frac{n}{2\, ^{3/2}} , \] has at most one stationary point. Splitting the integral at that point (if it exists) and applying the same argument on each interval shows that I⁻_n → 0.

Therefore \[ c_n = \frac{1}{4} \left( I_n^+ + I_n^{-} \right) \,\to\, 0 . \]

Another argument. The functions \[ \psi_n (u) = e^{-\mathbf{j}\,n/\sqrt{u}} \] are uniformly bounded and satisfy \[ \psi'_n (u) = \frac{\mathbf{j}\,n}{2\,u^{3/2}} \, \psi_n (u) . \] Since \( \displaystyle \quad \int_1^{\infty} \,\frac{\cos u}{u}\, {\text d}u \quad \) is a Dirichlet integral, one can invoke the uniform Dirichlet test (or the Jordan–Dirichlet theorem) to conclude immediately that cₙ → 0.

This example is quite interesting because it shows that the decay of Fourier coefficients can hold even when the classical Riemann–Lebesgue lemma is not applicable due to the lack of absolute integrability.    ■

End of Example 5

Example 6: One of the nicest applications is to high-frequency oscillatory Stieltjes integrals, where the oscillation is carried by the integrator rather than by the integrand. This is the exact analogue of the classical Riemann–Lebesgue lemma.

Example 6A: Let \[ F(t) = \sin t , \] which is 2π-periodic, of bounded variation on every bounded interval, and has mean zero. If 𝑓 ∈ 𝐵𝑉[𝑎, 𝑏], then \[ \int_a^b \, f(x)\,{\text d}\left( \sin nx \right) \,\to\,0 . \] Since \[ {\text d}\left( \sin nx \right) = n\,\cod \left( \sin nx \right) , \] this becomes \[ n\,\int_a^b f(x)\,\cos (nx)\,{\text d}x \,\to\, 0 . \] If er take 𝑓(𝑥) = 𝑥 or sin𝑥, we get \[ \int_{-\pi}^{\pi} \, x\,\cos (nx)\,{\text d}x = 0, \qquad \int_{-\pi}^{\pi} \, \sin x\,\cos (nx)\,{\text d}x = 0 . \]

At first glance this is surprising because of the factor n. The oscillations of cos⁡(nx) exactly compensate for it. For example, with 𝑓(𝑥) = 𝑥 or sin𝑥, \[ \int_{0}^{1} \, x\,\cos (nx)\,{\text d}x = \frac{n\,\sin n -1 + \cos n}{n}, \qquad \int_{0}^{1} \, \sin x\,\cos (nx)\,{\text d}x = \frac{n \left( \cos 1\,\cos n -1 \right) + n^2 \sin 1\,\sin n}{1 + n^2} . \] which does not tend to zero. This illustrates why one must be careful about the precise theorem: the endpoint term survives unless the theorem is stated with the appropriate normalization (or with 𝑓(𝑎) = 𝑓(𝑏) = 0.

However, if 𝑓(0) = 𝑓(1) = 0, say \[ f(x) = x\left( 1-x \right) , \] then \[ \int_0^1 \, x\left( 1-x \right) {\text d} \left( \sin (nx) \right) \,\to \, 0 . \]

Example 6B: Let \[ F(t) = \begin{cases} \phantom{-}1 , \qquad & 0 < t < \pi , \\ -1 , \qquad & \pi < t < 2\pi , \end{cases} \] be a square wave extended periodically.

Then F has bounded variation, period 2π2π, and mean zero. Its Stieltjes measure is \[ {\text d}F = -2 \,\sum_{k\in \mathbb{Z}} \ \delta_{(2k-1) \pi} + 2 \,\sum_{k\in \mathbb{Z}} \ \delta_{2k\, \pi} , \] so \[ \int_a^b \,f(x)\,{\text d} F(\pi x) \] is a weighted sum of the values of 𝑓 at the lattice points \[ x = \frac{k\pi}{n} . \] The theorem says that these rapidly oscillating alternating point masses produce complete cancellation: \[ \int_a^b \,f(x) \,{\text d} F(nx) \,\to \, 0 . \] Thus the theorem can be interpreted as an equidistribution result for alternating Dirac masses.

The integrator is a step function, so the integral reduces to a finite alternating sum: \[ \int_a^b \,f(x)\,{\text d} F(nx) = 2\,\sum_{2k\pi /n \in (a,b)} \ f \left( \frac{2k\pi}{n} \right) \] and the theorem asserts that this alternating sampling of 𝑓 tends to zero as the mesh π/n shrinks. This provides a concrete and distinctly Riemann–Stieltjes application that has no equally natural formulation in ordinary Riemann integration.

Example 6C: If F is periodic of bounded variation, it has a Fourier–Stieltjes expansion. The theorem implies that for every 𝑓 ∈ 𝐵𝑉[𝑎, 𝑏], \[ \int_a^b \,f(x)\,{\text d} F(nx) \,\to\, 0 , \] which is exactly the statement that the dilated Stieltjes measures dF(nx) converge weakly to zero.

This viewpoint appears in harmonic analysis and Example 6D: Let F be the sawtooth function \[ F(x) = x - \lfloow x \rfloor - \frac{1}{2} \]

Then F is periodic with period 1, has mean zero, and is of bounded variation. Moreover, \[ {\text d}f = {\text d}x - \sum_{k \in \mathbb{Z}} \ \delta_k , \] with δₖ is the Kronecker delta. So \[ {\text d}f = n\,{\text d}x - \sum_{k} \ \delta_{k/n} . \] Hence, \[ \int_0^1 \,f(x)\,{\text d}F(nx) = n\,\int_0^1 \,f(x)\,{\text d}x - \sum_{k=1}^{n-1} \, f \left( \frac{k}{n} \right) , \] up to endpoint conventions. The Riemann--Lebesgue theorem therefore yields \[ \sum_{k=1}^{n-1} \, f \left( \frac{k}{n} \right) = n\,\int_0^1 \,f(x)\,{\text d}x + o(1) , \] which is simply the convergence of Riemann sums written in an elegant Stieltjes form.

Example 6E: Suppose \[ F(x) = \sum_{m \in \mathbb{Z}} \ a_m \,H(x-m) , \] where H is the Heaviside function and the coefficients satisfy \[ a_{m+p} = a_m , \qquad \sum_{m=0}^{p-1} a_m = 0 . \] Then \[ {\text d} F = \sum_{m \in \mathbb{Z}} \ a_m \,\delta_m \] is a periodic signed measure with zero total mass. The Riemann-Lebesgue theorem gives \[ \sum_m \,a_m \,f\left( \frac{m}{n} \right) \, \to \. 0 . \] Thus every zero-average periodic finite-difference operator annihilates smooth trends asymptotically.

For example, \[ a_0 = 1, \qquad a_1 = -1 , \] gives \[ \sum_k \left[ f\left( \frac{2k}{n} \right) - f\left( \frac{2k+1}{n} \right) \right] \,\to \, 0 . \]

Example 6F: The first Bernoulli periodic function \[ B_1 (x) = \{ x \} - \frac{1}{2} , \qquad \{ x \} = x - \lfloor x \rfloor , \] is exactly the sawtooth function above.

The Euler--Maclaurin identity can be written \[ \sum_{k=0}^{n-1} \, f \left( \frac{k}{n} \right) = n\,\int_0^1 f(x)\,{\text d}x - \int_0^1 \, f(x)\,{\text d}B_1 . \] The Riemann--Lebesgue lemma immediately implies \[ \int_0^1 \, f(x)\,{\text d}B_1 = o(1) , \] recovering the leading term of Euler--Maclaurin without any differentiability assumptions on 𝑓.

It shows that the error term in approximating an integral by a Riemann sum is exactly a Riemann–Stieltjes integral against the periodic Bernoulli function, and the Riemann–Lebesgue lemma explains why this error tends to zero. This gives a conceptual bridge between oscillatory Stieltjes integration and one of the fundamental asymptotic formulas in analysis.    ■

Example 3:    ■

Example 9: We start with the power function p₃(x) = x³. Expanding it into Fourier series, we get

Example 11: ????????????? Let us consider the quadratic polynomial 𝑓(𝑥) = 𝑎 𝑥² + b 𝑥 + c, where 𝑎, 𝑏, and c are real constants. The Fourier coefficients are readily calculated as \begin{align*} a_0 &= \frac{1}{\ell}\,\int_{-\ell}^{+\ell} \,\left( a\,x^2 + b\,x + c \right) {\text d} x = 2\,c + \frac{2}{3}\,a\,\ell^2 , \\ a_n &= \frac{1}{\ell}\,\int_{-\ell}^{+\ell} \,\left( a\,x^2 + b\,x + c \right) \cos \left( n\,\frac{\pi}{\ell}\,x \right){\text d} x = \frac{4a\,\ell^2}{n^2 \pi^2}\, (-1)^n , \\ b_n &= \frac{1}{\ell}\,\int_{-\ell}^{+\ell} \,\left( a\,x^2 + b\,x + c \right) \sin \left( n\,\frac{\pi}{\ell}\,x \right){\text d} x = -\frac{2b\,\ell}{n\pi}\, (-1)^n . \end{align*}

1. Bary, N.K., A Treatise on Trigonometric Series, A Pergamon Press Book, New York, 1964.
2. Lebesgue, H., Sur les séries trigonométriques (On trigonometric series), Annales scientifiques de l'École Normale Supérieure, Série 3, Tome 20, 1903, pages: 453–485.