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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.)

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    are you saying that $(\Rightarrow)$ is false if $L^1$ means Lebesgue integrable functions?2012-03-22
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    Yes. In that case, the Riemann integral doesn't have to exist.2012-03-22
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    OK. but if I take $f=1_{\mathbb{Q}}$, won't I rather get both $f$ and $|f|$ to be integrable.2012-03-22
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    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.)2012-03-22
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    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$2012-03-22
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    Then the hint I gave in the answer applies. (Even in the Lebesgue integrable case, but I wouldn't call it an "improper *Riemann* integral". In this case, it's just a Lebesgue integral, not even an improper Lebesgue integral.)2012-03-22