I am stuck with the following problem:
$I$ is an interval of $\mathbb{R}$. If $f: I \rightarrow \mathbb{R}$ locally Riemann-integrable (i.e, its restriction to every compact interval is Riemann-integrable). Then $f$ is Lebesgue-integrable if and only if the improper Riemann-integral of $|f|$ exist, and in these cases, they coincide.
I think that Monotone Convergece Theorem may be useful if one approximates $f$ with $f \chi_{A_n}$ with ${A_n}$ a sequence of compact sets in $I$, but I am not sure about that. So I would like to know how to proceed.
Thanks in advance.
Assume without loss of generality that $I=(a,b)$ and consider $I_{n}=[a+1/n,b-1/n]$ for large $n$, assume that \begin{align*} (R)\int_{a}^{b}|f(x)|dx \end{align*} exists in improper Riemann sense, then \begin{align*} \infty>(R)\int_{a}^{b}|f(x)|dx=\lim_{n\rightarrow\infty}(R)\int_{I_{n}}|f(x)|dx, \end{align*} and so the Lebesgue integral of $|f|$ on $I$ that \begin{align*} (L)\int_{I}|f(x)|dx &=\lim_{n\rightarrow\infty}(L)\int_{I}|f(x)|\cdot 1_{I_n}dx\\ &=\lim_{n\rightarrow\infty}(L)\int_{I_{n}}|f(x)|dx\\ &=\lim_{n\rightarrow\infty}(R)\int_{I_{n}}|f(x)|dx\\ &=(R)\int_{a}^{b}|f(x)|dx, \end{align*} note that the first equality is by Monotone Convergence Theorem and also that for compact interval, a function is Riemann integrable then it is Lebesgue integrable and the integrals coincide.