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Is is true that if a function is Riemann integrable, then it is Lebesgue integrable with the same value? If it's true, how to prove it? If it's false, what is a counterexample?

Jean-Claude Arbaut
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unknown
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1 Answers1

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If you are talking about proper Riemann integrals, i.e. $f : [a,b] \to \Bbb{R}$ is bounded and the interval $[a,b]$ is compact, then this is true.

EDIT: In the following, all integrals $\int \dots \, dx$ are to be unterstood as the Riemann integral $\int_a^b \dots \, dx$. All integrals $\int \dots \, d\lambda(x)$ are to be understood as the Lebesgue integral $\int_{[a,b]} \dots \, \lambda(x)$.

For a proof, use that there are sequences $(\varphi_n)_n$ and $(\psi_n)_n$ of Riemann step functions such that $\varphi_n \leq f \leq \psi_n$ and

$$\int \varphi_n \, dx \to \int f \, dx \leftarrow \int \psi_n \, dx.$$

Depending on your exact definition of the Riemann integral, this is either a direct consequence, or an easy consequence of the definition.

By changing to $\max\{ \varphi_1, \dots, \varphi_n \}$ and $\min\{\psi_1, \dots, \psi_n\}$, we can assume w.l.o.g. that the sequences $(\varphi_n)_n$ and $(\psi_n)_n$ are increasing/decreasing.

On Riemann step functions $\gamma : [a,b] \to \Bbb{R}$, the Lebesgue integral and the Riemann integral coincide (why?). Hence,

$$ \int |\psi_n - \varphi_n| \, d\lambda(x) = \int \psi_n - \varphi_n \, dx \to \int f\, dx - \int f \, dx = 0. \qquad (\dagger) $$

By monotonicity, also $\varphi_n \to \varphi$ and $\psi_n \to \psi$ pointwise with $\varphi_1 \leq \varphi \leq f \leq \psi \leq \psi_1$.

By dominated convergence,

$$ \int |\psi_n - \varphi_n| \, d\lambda(x) \to \int |\psi - \varphi| \, d\lambda(x). $$

By $(\dagger)$, we get $\int|\psi- \varphi|\, d\lambda(x) = 0$ and hence $\psi = \varphi$ almost everywhere.

Because of $\varphi \leq f \leq \psi$, we get $f = \varphi = \psi$ almost everywhere, so that $f$ is Lebesgue measurable with

$$ \int f \, d\lambda(x) = \int \psi \, d \lambda(x) = \lim_n \int \psi_n \, d\lambda = \lim \int \psi_n \, dx = \int f \, dx. $$

This completes the proof.

For improper Riemann integrals, the claim is false however, as (cf. the answer by Peter) the example of $$\frac{\sin(x)}{x}$$ shows.

The point here is that Lebesgue integrability of $f$ implies integrability of $|f|$, whereas for the (improper) Riemann integral it can happen that $f$ is integrable, although $|f|$ is not.

Finally, one can even show (using a variant of the proof above) that a bounded function $f : [a,b] \to \Bbb{R}$ is Riemann integrable if and only if the set of discontinuities of $f$ is a set of Lebesgue measure zero.
PhoemueX
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  • What did you mean by "improper Riemann integrals"? – Bear and bunny Aug 03 '15 at 21:06
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    @Bearandbunny : Improper integrals are limits as the either or both of the bounds of integration approach something. If the integral $\displaystyle\int_{[0,\infty)} \frac{dx}{1+x^2}$ is viewed as a Lebesgue integral, then it is NOT improper, because it is not defined by first defining $\displaystyle\int_0^a \frac{dx}{1+x^2}$ and then taking a limit as $a\to\infty$. But $\displaystyle\int_0^\infty \frac{\sin x} x, dx$ can only be viewed as an improper integral because the integrals of both the positive and negative parts are infinite. ${}\qquad{}$ – Michael Hardy Aug 04 '15 at 03:18
  • @MichaelHardy: Learn a lot. Thanks. – Bear and bunny Aug 04 '15 at 04:05
  • What about improper Lebesgue integrals? – lalala Feb 27 '18 at 19:31
  • @lalala: Normally, one does not define these, since you then lose many of the nice properties of the Lebesgue integral. If you however define the improper Lebesgue integral as something like $\int_{a\ast}^{b\ast}f(t)d\lambda(t)=\lim_{x\to a, y\to b} \int_x^y f(t) d\lambda(t)$, where the integral inside the limit is a "proper Lebesgue integral" and where the stars are used to distinguish the improper integral, then Lebesgue and Riemann will again coincide (if both exist): The expression inside the limit is the same whether you use the Lebesgue or the Riemann integral. Thus, the limits coincide. – PhoemueX Feb 27 '18 at 20:16
  • Is the integral of $f(x)=\dfrac {\sin x}{x}$ not defined because $\int |f| = \infty$, or it it because $\int f^+ - \int f^-$ has the form $\infty - \infty$? – Ovi Mar 29 '18 at 05:21
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    @Ovi: Well, both :) depending on your exact definition of the Lebesgue integral. Some authors only define $\int f$ if $\int |f|<\infty$, and some authors only require at least one of $\int f_+$ or $\int f_-$ to be finite. In the case of $f(x)=\sin(x)/x$, neither condition is satisfied. – PhoemueX Mar 29 '18 at 05:31
  • Thanks! ${}{}{}$ – Ovi Mar 29 '18 at 05:35
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    @PhomueX aren't the conditions $\int |f| < + \infty$ and $\int f^+, \int f^- < + \infty$ equivalent? Because $|f| = f^+ + f^-$ and both $f^+,f^- \geq 0$ so $f^+,f^- \leq |f|$ and hence their integrals. – math maniac. Jul 02 '20 at 08:12
  • What I am trying to say is that $$\int |f| = \int f^+ + \int f^-$$ and since $\int f^+ , \int f^- \geq 0$ so $\int f^+, \int f^- \leq \int |f|.$ So $$\int |f| \lt + \infty\ \iff \int f^+, \int f^- \lt + \infty.$$ – math maniac. Jul 02 '20 at 08:20
  • @math.maniac: Yes, the two conditions are equivalent. But as I said, some authors only require that at least one of $\int f_+$ or $\int f_-$ to be finite in order for $\int f$ to be defined. – PhoemueX Jul 02 '20 at 10:18