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Is the following function Lebesgue measurable? $$f(x):= \frac{\sin x}{1 + x^2} $$The problem confuses me a bit, since it doesn't state where it wants it to be Lebesgue integrable. I figure that since $f$ is continuous on any interval $[a,b]$, further it is bounded by $1$, then it is Riemann-integrable (since the points where it is discontinuous is a Lebesgue null set), and since it is Riemann-integrable, it's definitely Lebesgue-integrable.

Is this correct? If yes, how do I extend "Lebesgue integrable on $[a,b]$" to just "Lebesgue integrable"?

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    $\begingroup$ Probably it means "Lebesgue integrable on $\mathbb R$". $\endgroup$ Sep 28 '15 at 13:33
  • $\begingroup$ Out of context one would expect the domain to be $\mathbb{R}$ (so that in fact the domain could be whatever measurable set you like). In this case it is Lebesgue integrable on $\mathbb{R}$, so that seems to support the conclusion. $\endgroup$
    – Ian
    Sep 28 '15 at 13:33
  • $\begingroup$ Are you asking about measurability or integrability? It is both measurable and Lebesgue integrable, anyway, since $|f|$ is a non-negative Lipschitz function bounded by $\frac{1}{1+x^2}$ that is clearly integrable over $\mathbb{R}$. $\endgroup$ Sep 28 '15 at 13:33
  • $\begingroup$ @JackD'Aurizio It's not nonnegative (maybe you meant $|f|$) $\endgroup$
    – Ian
    Sep 28 '15 at 13:34
  • $\begingroup$ @Ian: sure, fixed. Thanks. $\endgroup$ Sep 28 '15 at 13:36
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It is Lebesgue integrable on the entire line. Observe that $$\left| \frac{\sin x}{x^2 + 1} \right| \le \frac 1{x^2 + 1}$$ for all $x$. The integral of the latter function can be computed using Riemann integrals and the monotone convergence theorem.

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For all $x\in\mathbb R$, $$0\leq \left|\frac{\sin(x)}{1+x^2}\right|\leq \left|\frac{1}{1+x^2}\right|=\frac{1}{1+x^2}.$$

Since $x\mapsto \frac{1}{1+x^2}$ is Lebesgue integrable (even Riemann integrable), $x\mapsto \frac{\sin(x)}{1+x^2}$ is Lebesgue integrable (and even Riemann integrable).

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  • $\begingroup$ Why is $x \mapsto \frac{1}{1 + x^2}$ Lebesgue integrable? You should know I have yet to meet a single Lebesgue-integrable function in my studies other than the very trivial ones. $\endgroup$
    – Adaman
    Sep 28 '15 at 13:36
  • $\begingroup$ Because $x\mapsto \frac{1}{1+x^2}$ is Riemann integrable and Riemann integrable $\implies $ Lebesgue integrable ! $\endgroup$
    – Surb
    Sep 28 '15 at 13:37
  • $\begingroup$ ... allowing me to conclude it is Lebesgue integrable on every $[a,b]$, which is the only theorem I can refer to. Does it also hold for the entire real line? $\endgroup$
    – Adaman
    Sep 28 '15 at 13:38
  • $\begingroup$ I don't understand your question. If it's integrable on $\mathbb R$, it's of course integrable on all $[a,b]$. $\endgroup$
    – Surb
    Sep 28 '15 at 13:41
  • $\begingroup$ No, I meant that using the theorems at $\textbf{my}$ disposal, I can only conclude it is integrable on $[a,b]$ (which doesn't imply integrability on all of $\mathbb{R}$). So what I'm asking is, does Riemann-integrability on all of $\mathbb{R}$ imply Lebesgue integrability on all of $\mathbb{R}$, and the theorem in my book just happens to not state that? (Obviously that's what your answer implies, I'm just curious why it's not mentioned in my book?) $\endgroup$
    – Adaman
    Sep 28 '15 at 13:42
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You may actually prove: $$ I=\int_{0}^{+\infty}\frac{\sin x}{x^2+1}\,dx \leq \sqrt{\frac{2}{3}\,\log 2}\tag{1}$$ by using integration by parts and the Cauchy-Schwarz inequality:

$$ I = 2\int_{0}^{+\infty}\frac{s(1-\cos s)}{(1+s^2)^2}\,ds\leq 2\sqrt{\int_{0}^{+\infty}\frac{(1-\cos(s))^2}{s^3}\,ds\cdot \int_{0}^{+\infty}\frac{s^5}{(s^2+1)^4}\,ds}.\tag{2}$$

On the other hand, $\int_{0}^{+\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}$ and $\left|\sin x\right|\leq 1$. Moreover, $\frac{\sin x}{x^2+1}$ is an improperly Riemann-integrable function over $\mathbb{R}^+$ by Dirichlet's test, since $\sin x$ has a bounded primitive and $\frac{1}{x^2+1}$ is a decreasing function with limit zero.

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