# f Lebesgue integrable iff improper Riemann integral of |f| converges

Let $f:[a,b)\rightarrow \mathbb{R}$ be a continuous function with $\lim_{x \uparrow b} f(x) = \infty$. I have to show that $f$ is Lebesgue integrable if and only if the improper Riemann integral of |f| ($\lim_{\beta \to b} \int_{a}^\beta |f(x)|\,dx$) converges.

Since $f$ is measurable I know that $f$ being Lebesgue integrable is equal to $\int_{[a,b)} |f(x)|\,d\lambda < \infty$.

Any help/tip would be greatly appreciated!

• Use the dominated convergence theorem. – Lord Shark the Unknown Dec 18 '17 at 18:50

For the restated problem: This will follow from

1. If $f$ is Riemann integrable on $[c,d]$ then $f$ is Lebesgue integrable on $[c,d],$ and the two integrals agree.

2. If $f$ is Lebesgue integrable on $[a,b],$ then

$$\lim_{\beta \to b^-}\int_a^\beta f(x)\,dx = \int_a^b f(x)\,dx.$$

To prove 2., note it suffices to treat a sequence $\beta_n \to b^-.$ Set $f_n = \chi_{[a,\beta_n]}\cdot f$ and apply the dominated convergence theorem.

This is false:

$$\int_0^1 \frac{\sin(1/(1-x))}{1-x}\, dx$$

converges as an improper Riemann integral, although

$$\int_0^1\left | \frac{\sin(1/(1-x))}{1-x}\right |\, dx= \infty.$$

• True, I forgot to add absolute value to the second part of theorem, sorry about that! Could you check again, and help me with proving this version? – Metod Jazbec Dec 18 '17 at 19:17

The only if part is true. For $\beta<b$, $|\chi_{[a,\beta]}(x)f(x)|\leq|f(x)|$ and $|f|$ is Lebesgue integrable, so by Lebesgue Dominated Convergence Theorem we have \begin{align*} \lim_{\beta\rightarrow b}\int_{[a,\beta]}|f(x)|dx=\int_{[a,b)}|f(x)|dx \end{align*} in the sense of Lebesgue integrals. As $f$ is continuous, then \begin{align*} \int_{[a,\beta]}|f(x)|dx=\int_{a}^{\beta}|f(x)|dx, \end{align*} where the right-sided is Riemann integral.

For the other direction, you may use Monotone Convergence Theorem, put $\beta_{n}=b-1/n$ for large $n$.