Is $f(x)=\sin x$ integrable?

Question

I'd like to know if the following reasoning to prove that $f(x)=\sin x$ is not integrable is correct, or what's the mistake I'm making.

Consider the following definition and Corollary taken from Folland's Real Analaysis:

Definition: Consider a measure space $(X,M, \mu)$. If $f:X \to \Bbb R$, we say that $f$ is integrable if both $\int_X f^+$ and $\int_X f^-$ are finite. It is clear that $f$ is integrable iff $\int_X |f|<\infty$ since $|f|=f^+ + f^-$.

Corollary 2.2: If $X$ and $Y$ are topological spaces, every continuous $f:X\to Y$ is $(B_X, B_Y)$-measurable, where $B_X$ and $B_Y$ are the Borel $\sigma$-aglebras on $X$ and $Y$, respectively.

Now, the function $f:\Bbb R \to \Bbb R$ defined by $f(x)=\sin x$ is continuous, and by the corollary, it's Borel-measurable.

We have that $\int_{\Bbb R} f = \int_{-\infty}^\infty \sin (x) dx = 0$, but $\int_{\Bbb R} f^+= \int_{\Bbb R} f^-=\infty$, so, $f(x)=\sin x$ is not integrable.

Note: I'm doing this because I'm trying to understand why do we have $\int_X |f|<\infty$ and not $\int_X f<\infty$ in the definition of integrable function, and I'm trying to find a counterexample that $\int_X f<\infty$ doesn't imply $\int_X |f|<\infty$. If someone has a valid counterexample it would be nice to know it.

It is not clear why the fact that $\lim_{M \to \infty} \int_{-M}^M \sin(x) , dx = 0$ (which, btw, is not the same as $\int_{\mathbb{R}} f = 0$!) is relevant. If you can argue that $\int_{\mathbb{R}} f^{+} = \infty$ then clearly $f$ is not integrable according to the definition you wrote (as $\int_{\mathbb{R}} f^{+}$ is not finite). — levap, Jun 21 '20 at 22:28
@Alex $\int_{\Bbb R} f = 0$ is proved here https://math.stackexchange.com/questions/725025/is-int-infty-infty-sin-x-mathrmdx-divergent-or-convergent — Twnk, Jun 21 '20 at 22:31
@Twink: In the general setting of measure theory, the integral $\int_X f$ is defined in terms of $\int_{X} f^{+}$ and $\int_X f^{-}$ and you require both of them to be finite which then implies that also $\int_X |f|$ exists and is finite. It is a legitimate question to ask why this is done. How would you define $\int_X f$ for a general non-negative $f$ on a general measure space $X$? — levap, Jun 21 '20 at 22:33
@Twink: https://www.wolframalpha.com/input/?i=integral%28sinx%2C+x%3D-inf%2C+%2Binf%29 — Alex, Jun 21 '20 at 22:34
@levap It's defined as $\int f = \int f^+ - \int f^-$, but Folland also says that $\infty - \infty$ is not well defined. — Twnk, Jun 21 '20 at 22:35
@Twink The post you liked says that the Cauchy principal value of $\int_{-\infty}^\infty\sin x,dx$ is $0$, not that $\int_{\Bbb R}\sin x,dx=0$. — , Jun 21 '20 at 22:49

Mark · Accepted Answer · 2020-06-21T22:45:03.320

1

First of all, $\int_{-\infty}^{\infty} \sin(x)dx=0$ is false. Even if we consider the improper Riemann integral, it is not defined as $\lim_{R\to\infty}\int_{-R}^R f(x)dx$, but as $\lim_{R,M\to\infty}\int_{-M}^R f(x)dx$. So even as an improper Riemann integral $\int_{-\infty}^{\infty} \sin(x)dx$ is divergent.

Anyway, we are discussing the Lebesgue integral here. As you noted correctly the integral is divergent, for example because $f^{+}$ is not integrable.

I'll write a few words about your note. You ask why $\int_X f<\infty$ is not the definition of an integrable function. Well, actually it is, but first of all we have to define what does $\int_X f$ even means for a general measurable function. Thing is, Lebesgue integration is defined in a few steps. Usually it is first defined for non negative measurable functions. After we did that we can define it for a general measurable function $f$: it is called integrable if both integrals $\int_X f^{+}$ and $\int_X f^{-}$ are finite, and then we define $\int_X f=\int_X f^{+}-\int_X f^{-}$. But it is also easy to see that the condition $\int_X f^{+},\int_X f^{-}<\infty$ is equivalent to the condition $\int_X |f|<\infty$ (remember, we already defined the Lebesgue integral for non-negative functions), so we get an equivalent definition of $f$ being integrable that way.

In other words, it is true that $\int_X f$ is a finite number if and only if $\int_X |f|$ is.

edited Jun 21 '20 at 22:45

answered Jun 21 '20 at 22:37

Mark

39,605

Ok. The inequality $|\int f | \leq \int |f|$ was making me think that only the implication $\int |f| < \infty \implies \int f < \infty$ was valid. So, if I need to prove that a function is integrable, is it enought to prove that $\int f < \infty$? Even if $f$ takes negative values? – Twnk Jun 21 '20 at 23:54
I wouldn't write it $\int f<\infty$, because if the function is not non-negative then this integral can be also equal $-\infty$ or might be undefined. It is more correct to say that it is enough to show that the integral $\int f$ is a real number. However, in order to do that you need to know what this integral equals to, and most of the time we don't know it. Also, remember it has to be the Lebesgue integral. For example, the function $\frac{\sin(x)}{x}$ has a conditionally convergent Riemann improper integral, but the Lebesgue integral simply doesn't exist. – Mark Jun 22 '20 at 07:10
But then is it enough to prove that $|\int f| < \infty$ if I need to prove that $f$ is integrable? Which means $|\int f| < \infty \implies \int |f| < \infty$. – Twnk Jun 22 '20 at 21:29
Yes, it is enough to show that $|\int f|<\infty$. But again, $\int f$ has to be the Lebesgue integral, not other types of integrals. (I gave an example with $\frac{\sin(x)}{x}$ is my previous comment). In most cases it is much easier to show that $\int |f|<\infty$, because it is easier to work with non negative functions. – Mark Jun 22 '20 at 21:56
But what about the theorem that says that if $f$ is bounded on $[a,b]$ and it's Riemann integrable, then it is Lebesgue integrable, and the two integrals coincide? – Twnk Jun 22 '20 at 22:03
Doesn't it work for that example since $\frac{\sin x}{x}$ is bounded on any closed interval? – Twnk Jun 22 '20 at 22:09
Yes, if a function is Riemann integrable then it is Lebesgue integrable and the integrals are equal. It is also true for functions which have an absolutely convergent improper integral. However, it is not true for functions which have a conditionally convergent improper integral. In Lebesgue integration there is no such thing as conditional convergence, we always have that $f$ is integrable if and only if $|f|$ is. So $\frac{\sin x}{x}$ is indeed Lebesgue integrable on any bounded interval. But on the whole real line it isn't, because $\int_\mathbb{R}|\frac{\sin x}{x}|dx=\infty$. – Mark Jun 22 '20 at 23:18
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Thanks. I'm gonna have to learn about this topic of improper integrals. – Twnk Jun 22 '20 at 23:35

score 1 · Answer 2 · answered Jun 21 '20 at 22:43

You have a few definitions mixed up. For any function $f:X\to\mathbb{R}$, the value $\int_{X} f dx = \int_{X}f^+ dx - \int_{X}f^- dx$ is defined only if $ \int_{X}f^+ dx , \int_{X}f^- dx < \infty$. Since you noted that $ \int_{X}f^+ dx = \int_{X}f^- dx = \infty$, the value $\int_{X} dx$ is simply not defined.

The integral $\int_{-\infty}^\infty f(x)dx$ is a Riemann integral and is defined only if the following limits exist $\lim_{m,n\to\infty}=\int_{-m}^n f(x)dx$. What you have noticed is that $\lim_{M\to\infty}\int_{-M}^Mf(x)dx = 0$. But we do not always have $\lim_{M\to\infty}\int_{-M}^Mf(x)dx = \lim_{m,n\to\infty}\int_{-m}^n f(x)dx$

Is $f(x)=\sin x$ integrable?

2 Answers2