These is the definition I use:

Definition 1: An improper integral will be an integral of the form: $$\int_a^b f(x) \, dx $$ Where $f$ is Riemann Integrable, on every finite subinterval, $[s,t] \subseteq (a,b)$. ($a,b$ can be $ \pm \infty$). If for some $c \in (a,b)$, the limits exists, $$ \int_a^b f(x) \, dx := \lim_{s \rightarrow a } \int_s^c f (x) \, dx + \lim _{ t \rightarrow b} \int_c^t f(x) \, dx . $$

An improper integral is \textit{absolutely convergent} if $\int_a^b |f(x)| \,dx $ converges.

Question: I know if $f$ is Riemann Integrable on $[a,b]$ then (i) it is measurable, and (ii) its integral conincides with the Lesbgue Integral. But if $f$ is improperly Riemann Integrable, then is it even measurable in the first place?


Since $f$ and, hence, $|f|$ is Riemann integrable on $[a + 1/n,b-1/n]$, for every positive integer $n$ the restriction of $|f|$ to any such interval is measurable.

Hence, for every $\alpha \in \mathbb{R}$, the set $E_{\alpha,n}=\{x \in [a+1/n,b-1/n]:|f(x)| > \alpha \}$ is measurable. Consequently, for every $\alpha$ we have measurability of

$$E_\alpha = \{x \in [a,b]:|f(x)| > \alpha \} = \bigcup_{n=1}^\infty E_{\alpha,n} \cup F,$$

since $F$ is either $\phi$, $\{a\}$, $\{b\}$, or $\{a,b\}$ which is a measure-zero measurable set and, hence, $E_\alpha$ is a countable union of measurable sets.

Therfore, $f$ is measurable on $[a,b]$.

Also, the Lebesgue integral of $|f|$ coincides with the improper Riemann integral of $|f|$. For one side (and similarly for the other) we have by the monotone convergence theorem,

$$\int_c^b |f(x)| \, dx = \lim_{t \to b-} \int_c^t |f(x)| \, dx = \lim_{t \to b-} \int_{[c,t]} |f| = \lim_{t \to b-} \int_{[c,b]} |f| \chi_{[c,b]} = \int_{[c,b]}|f|$$

Here I use $\int_c^d g(x) \, dx$ to denote a Riemann integral and $\int_{[c,d]} g$ to denote a Lebesgue integral.

Now, the Lebesgue and improper Riemann integrals of $f$ can be shown equivalent using the dominated convergence theorem.

  • $\begingroup$ Thanks a lot , for $E_{\alpha,n}$ why is there an absolute value "$|f|>\alpha$? If $f$ is only improperly Riemann Integrable and not absolutely ( so we cannot apply DCT), do the integrals still match? $\endgroup$ – Bryan Shih Jul 11 '17 at 22:25
  • $\begingroup$ @CWL: I made an edit. In the first part I am showing that $|f|$ is measurable. Since $f$ is Riemann integrable on any subinterval of $[a,b]$, then so is $|f|$. For your second question, if $f$ is improperly Riemann integrable and $|f|$ is not, then the Lebesgue integral $\int_{[a,b]} f$ need not exist. An example is $f(x) = (1/x) \sin(1/x)$ on $[0,1]$. $\endgroup$ – RRL Jul 11 '17 at 22:41
  • $\begingroup$ Wait, i thought the integral of $\int_{[a,b]} f $ always exists. At least in my definition, when we take the Lebesgue measure, it exists for all Lebesgue measurable functions. My definition of "lebesgue integrable" is when $\int |f| < \infty$. I don't see how the example works :( $\endgroup$ – Bryan Shih Jul 12 '17 at 8:15
  • 1
    $\begingroup$ The function $f(x) = (1/x)sin(1/x)$ is the standard textbook example of a function which is improperly Riemann integrable but not Lebesgue integrable on $[0,1]$. Precisely as you say because $\int_{[0,1]} |f| = +\infty$ but the improper integral $\int_0^1 f(x) \, dx$ converges in this case. $\endgroup$ – RRL Jul 12 '17 at 8:40
  • 1
    $\begingroup$ Changing variables with $y = 1/x$ we get $\int_0^1 (1/x) \sin(1/x) \, dx = \int_1^\infty \frac{\sin y}{y} \, dy < \int_0^\infty \frac{\sin y}{y} \, dy = \frac{\pi}{2}.$ I'm sure you know that one. However, $\int_1^\infty \frac{|\sin y|}{y} \, dy = +\infty$. It diverges as an improper Rieman integral and fails to exist (is infinite) as a Lebesgue integral. Proving this last result comes up on this site very frequently. $\endgroup$ – RRL Jul 12 '17 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.