Proof verification: existence of improper integral, given that the integral of the absolute value is finite.

Question

I'm working on the following problem, but I don't know if my proof is correct, so I'm looking for some feedback. Also, my approach seems quite tedious, so I'm curious whether anyone can think of a simpler proof. Thanks in advance!

Suppose $f$ is continuous on $(a,b)$ and $\int_a^b |f(x)| \mathrm{d}x < \infty$; $a$ can be $-\infty$ and $b$ can be $+\infty$. Show the integral $\int_b^a f(x)\mathrm{d}x$ exists and is finite.

So I tried the following: $$0\leq\left|\int_a^bf(x)\mathrm{d}x\right|\leq\int_a^b|f(x)|\mathrm{d}x<\infty,$$ so $\left|\int_a^bf(x)\mathrm{d}x\right|\leq M$, for an $M\in\mathbb{R}$, given that $\int_a^bf(x)\mathrm{d}x$ exists. Thus in that case $\int_a^bf(x)\mathrm{d}x$ is indeed finite. (i)

What's left is to proof that $\int_a^bf(x)\mathrm{d}x$ indeed exists.

In the case that the integral is a 'normal' (that is non-improper) Riemann integral, the continuity of $f$ directly implies the existence of $\int_a^bf(x)\mathrm{d}x$. So we consider the case the the integral is improper. I assume that for every $c\in(a,b)$, $f$ is integrable on $[a,c]$. The case where either $a=-\infty$ or $f$ is non-integrable on $[a,c]$ follows analogously. If both the following case and the case that either $a=-\infty$ or $f$ is non-integrable on $[a,c]$ occur, then apply the following argument twice.

So $b$ is either $\infty$ (ii) or $f$ is not integrable on $(a,b]$ (iii).

First assume that $b=\infty$. We know that $\int_a^b|f(x)|\mathrm{d}x<\infty$, so $$\lim_{d\to\infty}\int_a^d|f(x)|\mathrm{d}x=L,$$ for an $L\in\mathbb{R}$. Thus, given $\epsilon>0$ for certain $y>0$, $d>y$ implies $$\left|\int_a^d|f(x)|\mathrm{d}x-L\right|<\epsilon\\\left|\int_a^d|f(x)|\mathrm{d}x-\int_a^{\infty}|f(x)|\mathrm{d}x\right|=\left|\int_d^{\infty}|f(x)|\mathrm{d}x\right|<\epsilon$$ So from here, there exists a partiton $P$ such that $M(f,P)-m(f,P)<\epsilon$, or $\int_a^{d}f(x)\mathrm{d}x+(M(f,P_{(d,\infty)})-m(f,P_{(d,\infty)}))<\int_a^{d}f(x)\mathrm{d}x+\epsilon$, and thus the integral exists.

Now, since $f$ is continuous on $(a,b)$, we know that $\int_a^df(x)\mathrm{d}x$ exists. Thus we conclude that $\int_a^bf(x)\mathrm{d}x=\int_a^df(x)\mathrm{d}x+\int_d^{\infty}f(x)\mathrm{d}x<\int_a^df(x)\mathrm{d}x+\epsilon$. Thus $$\left|\int_a^bf(x)\mathrm{d}x-\int_a^df(x)\mathrm{d}x\right|<\epsilon,$$ and we can conclude that $\lim_{d\to\infty}\int_a^df(x)\mathrm{d}x=\int_a^{\infty}f(x)\mathrm{d}x$ exists. (ii)

Now assume that $f$ is not integrable on $(a,b]$, with $b<\infty$. Again, $\int_a^b|f(x)|\mathrm{d}x$ exists and is finite, say $\int_a^b|f(x)|\mathrm{d}x=L$, for an $L\in\mathbb{R}$. Thus $\int_a^b|f(x)|\mathrm{d}x=\lim_{d\to b^-}\int_a^d|f(x)|\mathrm{d}x=L$. Thus given $\epsilon>0$, $\exists\delta>0$ such that $$\left|\int_a^{b-\delta}|f(x)|\mathrm{d}x-L\right|<\epsilon\\\left|\int_a^{b-\delta}|f(x)|\mathrm{d}x-\int_a^{b}|f(x)|\mathrm{d}x\right|=\left|\int_{b-\delta}^{b}|f(x)|\mathrm{d}x\right|<\epsilon$$

So from here, there exists a partiton $P$ such that $M(f,P)-m(f,P)<\epsilon$, or $\int_a^{b-\delta}f(x)\mathrm{d}x+(M(f,P_{(d,b-\delta)})-m(f,P_{(d,b-\delta)}))<\int_a^{b-\delta}f(x)\mathrm{d}x+\epsilon$, and thus the integral exists.

From this we conclude that $$\left|\int_{a}^{b}f(x)\mathrm{d}x-\int_{a}^{b-\delta}f(x)\mathrm{d}x\right|<\epsilon,$$ and thus that $\lim_{d\to b^-}\int_{a}^{d}f(x)\mathrm{d}x$ exists. (iii)

From (ii) and (iii) we conclude that $\int_{a}^{b}f(x)\mathrm{d}x$ exists. The case that the integral $\int_{a}^{c}f(x)\mathrm{d}x$ for $c\in(a,b)$ is also improper follows analogously. From (i), we now coclude that $\int_{a}^{b}f(x)\mathrm{d}x$ is finite.

Answer 1

The notion of improper Riemann integral is introduced precisely because the Riemann integral is not defined for unbounded functions and/or unbounded intervals. How are the upper and lower sums, $M(f,P)$ and $m(f,P)$, defined when the partition supposedly covers the unbounded interval $[d,\infty)$? There are special cases where infinite partitions and corresponding sums can be manipulated in a similar fashion to what is done for the true Riemann integral, but in general the machinery breaks down.

Focusing just on the integral over the semi-infinite interval $[a,\infty)$, if $f$ is Riemann integrable over $[a,c]$ for all $c > a$ then the improper integral is defined (pending existence) as

$$\int_a^\infty f(x) \, dx = \lim_{c \to \infty} \int_a^c f(x) \, dx.$$

Given that $\int_a^\infty |f(x)| \, dx$ exists, we have to prove that $\int_a^\infty f(x) \, dx$ exists. It is tempting to try immediately to show that the tail $\int_d^\infty f(x) \, dx$ can be made arbitrarily small by choosing sufficiently large d, but this is circular in that we have not yet established that an improper integral of $f$ over $[d,\infty)$ exists.

As with sequences and series, proof of existence without a prior candidate for the limit can be facilitated by a Cauchy criterion:

The improper integral exists if and only if for any $\epsilon > 0$ there exists $K > 0$ such that for all $c_2 > c_1 \geqslant K$ we have

$$\left|\int_{c_1}^{c_2}f(x) \, dx \right| < \epsilon$$

Armed with this theorem, the proof in question is straightforward.

Since the improper integral of $|f|$ exists, given $\epsilon > 0$ there exists $K$ such that with $c_2 > c_1 > K$ we have

$$ \tag{1} \left| \int_{c_1}^{c_2} f(x) \, dx\right| \leqslant \int_{c_1}^{c_2} |f(x)| \, dx < \epsilon.$$

Therefore, the improper integral of $f$ exists. Note that the first inequality in (1) is a well-known result for Riemann integrals.

Proof of Cauchy criterion:

Proof of the forward implication is straightforward. We are using the reverse implication here and the proof is as follows.

Define the sequence $I_n = \int_a^{a +n} f(x) \, dx $. From the hypotheses, given $\epsilon > 0$ there exists $K$ such that if the positive integers $m$ and $n$ satisfy $m > n \geqslant K- a$, then $a+m > a+n > K$ and

$$\tag{2}|I_m - I_n| = \left|\int_{a+n}^{a+m} f(x) \, dx \right| < \epsilon/2.$$

Hence, $(I_n)$ is a Cauchy sequence of real numbers and therefore converges to some real number $I$:

$$\tag{3}\lim_{n \to \infty} \int_a^{a+n} f(x) \, dx = I$$.

We can write

$$\tag{4}\left| \int_a^c f(x) \, dx - I \right| = \left| \int_{a+n}^c f(x) \, dx + \int_{a}^{a+n} f(x) \, dx - I \right| \\ \leqslant \left| \int_{a+n}^c f(x) \, dx \right| + \left| \int_{a}^{a+n} f(x) \, dx - I \right| .$$

If $c$ is sufficiently large, we can find $n$ sufficiently large with $c > a+n$, such that, using (2) and (3), the terms on the RHS of (4) are each less than $\epsilon/2$ . Therefore, the improper integral exists and takes the value $I$.