# Show that is $\int_A f$ exists, then so does $\int_A |f|$

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.

• What is $A$? This result seems incorrect, $\int_0^{+\infty} \frac{sin x}{x}$ would be a counter-example. – Jack Nov 22 '15 at 21:06
• $A \subset E^n$ Also, assume $A$ is closed interval to simplify the problem. – Meecolm Nov 22 '15 at 21:07
• @Jack I think $A$ is bounded (it should be otherwise this is wrong). – Hamed Nov 22 '15 at 21:08
• @Hamed Yes I believe you are correct, even though the problem statement is exactly as stated above. – Meecolm Nov 22 '15 at 21:09
• 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 $-|f|\leqslant f\leqslant|f|$. – Did Nov 22 '15 at 21:30

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