Let $f: [a,b] \to [0,\infty)$ and $f$ is Riemann Integrable on every subinterval $[a + \epsilon,b]$ for $\epsilon > 0$. Suppose that the improper Riemann integral exists. That is
$$I = \lim_{\epsilon \to 0} \int_{a + \epsilon}^{b} f(x) dx < \infty$$
exists. Prove that $f$ is Lebesgue integrable on $[a,b]$ and that $\int_{[a,b]} f(x) dx = I$.
I found another posting of this problem written in the same way, but it used dominated convergence theorem, which I have not covered. The others I saw were written slightly differently and/or weren't making too much sense.
Also, maybe I'm not seeing something clearly, but here's something I thought:
The improper integral I is finite (it exists). If the improper integral exists, isn't it equal to the "proper" integral? That would mean the "normal" proof of showing that every Riemann Integrable function is Lebesgue integrable would apply. However, I have a feeling this is not the way to prove it or else this question wouldn't be asked.
Thanks for the help!
Best Answer
Since $f$ is nonnegative, the Lebesgue integral must exist (but may be infinite). With the existence of the improper Riemann integral we can show that the Lebesgue integral is finite and $f$ is "Lebesgue integrable" on $[a,b]$.
For all sufficiently large $n$ we have $[a+1/n,b] \subset [a,b]$. Since $f \chi_{[a+1/n,b]} \nearrow f$ as $n \to \infty$, it follows by the monotone convergence theorem that
$$\int_{[a,b]} f = \int_{[a,b]} \lim_{n \to \infty}f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{[a,b]} f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{a+ 1/n}^b f(x) \, dx = \int_a^bf(x) \, dx < +\infty$$
Here we have applied the equivalence of the Lebesgue and Riemann integrals on the interval $[a+1/n,b]$