# Does Absolute Improper Integrability imply Lebesgue Integrability?

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.

• 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? – Cy L Shih Jul 11 '17 at 22:25
• @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 $[0,1]$. – RRL Jul 11 '17 at 22:41
• 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 :( – Cy L Shih Jul 12 '17 at 8:15
• 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. – RRL Jul 12 '17 at 8:40
• 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. – RRL Jul 12 '17 at 8:43