Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.)

share|improve this answer
    
are you saying that $(\Rightarrow)$ is false if $L^1$ means Lebesgue integrable functions? –  Kuku Mar 22 '12 at 6:46
    
Yes. In that case, the Riemann integral doesn't have to exist. –  mrf Mar 22 '12 at 6:52
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.