# Improper Riemann integral

I wish to prove or disprove the following statement:

$f\in L^1[0,\infty]$ if and only then $f$ has an improper Riemann integral on $[0,\infty)$.

I think $(\Leftarrow)$ is false. If we let $f(x) = \frac{\sin x}{x},~x\gt 0$, then $$\int_0^\infty \frac{\sin x}{x}~dx = \frac{\pi}{2},$$

but $$\int_0^\infty \left|\frac{\sin x}{x}\right|~dx = \infty.$$ So $f\notin L^1[0,\infty)$.

How about $(\Rightarrow)$? I can't seem to think of any counterexample, and I don't see how to show that it is true.

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I assume that by $L^1$ you mean Lebesgue integrable functions with $\int_0^\infty |f|\,dx < \infty$? If so, take $f$ as the characteristic function of $\mathbb{Q}$.
(If you mean Riemann integrable functions, the statement is true. Let $f_+ = \max\{f, 0\}$ and $f_- = -\min\{f, 0\}$. Then $f = f_+ - f_-$. Show that $\int_0^\infty f_+\, dx < \infty$ and $\int_0^\infty f_+\, dx < \infty$ using the triangle inequality.)
are you saying that $(\Rightarrow)$ is false if $L^1$ means Lebesgue integrable functions? –  Kuku Mar 22 '12 at 6:46
OK. but if I take $f=1_{\mathbb{Q}}$, won't I rather get both $f$ and $|f|$ to be integrable. –  Kuku Mar 22 '12 at 7:05
Then $f$ is Lebesgue integrable, but not Riemann integrable. (You rarely talk about improper Lebesgue integrals, and since you explicitly wrote improper Riemann integral, that's how I interpreted it.) –  mrf Mar 22 '12 at 7:10
Thanks. What I hope to find is a function such that $\int_0^\infty |f|~ dx \lt \infty$ but $\int_0^\infty f ~dx =\infty$ –  Kuku Mar 22 '12 at 7:24