1
$\begingroup$

Let $f$ be a real-valued function on $A$ in $E^n$. Show that is $\int_A f$ exists, then so does $\int_A |f|$ and $\left | \int_A f\right | \le \int_A |f|$.

I am trying to finish Rosenlicht's Introduction to Analysis and the last chapter is awful. But I do want to finish it because I have come this far. Could someone show me how to do this, so I can at least have some example problems to understand the material. This chapter (Multiple Integrals) really just left more confused than before I started reading the chapter.

Thank you in advance.

$\endgroup$
  • $\begingroup$ What is $A$? This result seems incorrect, $\int_0^{+\infty} \frac{sin x}{x}$ would be a counter-example. $\endgroup$ – Jack Nov 22 '15 at 21:06
  • $\begingroup$ $A \subset E^n$ Also, assume $A$ is closed interval to simplify the problem. $\endgroup$ – Meecolm Nov 22 '15 at 21:07
  • $\begingroup$ @Jack I think $A$ is bounded (it should be otherwise this is wrong). $\endgroup$ – Hamed Nov 22 '15 at 21:08
  • $\begingroup$ @Hamed Yes I believe you are correct, even though the problem statement is exactly as stated above. $\endgroup$ – Meecolm Nov 22 '15 at 21:09
  • 1
    $\begingroup$ By definition, Riemann integrable functions are bounded. To show the first part, note that for every partition $P$ and every cell $c$ in $P$, $\sup\limits_c|f|-\inf\limits_c|f|\leqslant\sup\limits_cf-\inf\limits_cf$ hence $U(|f|,P)-L(|f|,P)\leqslant U(f,P)-L(f,P)$, which shows that $|f|$ is Riemann integrable. For the second part, note that $-|f|\leqslant f\leqslant|f|$. $\endgroup$ – Did Nov 22 '15 at 21:30
0
$\begingroup$

In general, the statement is false: $\int_0^\infty {{\sin x}\over{x}} dx$ is well-defined by the alternating series theorem, but $\int_0^\infty |{{\sin x}\over{x}}| dx$ is not.

The statement is true if $A$ is closed and bounded (hence compact). In that case, any continuous measure $\mu$ on $A$ is finite, and so $\int_A f d\mu \leq \infty$ if and only if $f$ is finite a.e. with respect to $\mu$. If $f$ is a.e. finite, then so is $|f|$, from which the first part of the result follows.

The second part follows from Jensen's inequality. $|\cdot|$ is convex on the real line, and hence

$$ {\biggl |}\int_A f d\mu \biggr | \leq \int_A |f| d\mu $$

$\endgroup$
  • $\begingroup$ I realize that I gave an over-general answer, I guess. Your problem concerns only continuous functions and Riemann integration, so of worrying about a general continuous measure, you can simply consider Lebesgue measure, which corresponds to Riemann integration. The integral is well-defined only if the function is bounded, and the rest goes through as before. (Edited to correct a misspelling.) $\endgroup$ – JWLM Jan 29 '17 at 6:35

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.