I have the following proposition:
Proposition: Let $-\infty<a<b\leq\infty$ and $f:[a,b]\rightarrow[0,\infty]$ such that
$$\lim_{x\ \rightarrow\ b}R\int_{a}^{x}f(t)\ dt<\infty$$
Where "$R\int$" means integration in "Riemann sense". Then $f$ is Lebesgue integrable, i.e. $\int_{[a,b]}f\ d\mu<\infty$ where $\mu$ is the Lebesgue measure, and
$$\lim_{x\ \rightarrow\ b}R\int_{a}^{x}f(t)\ dt = \int_{[a,b]}f\ d\mu$$
Idea: I'm thinking in consider two cases: First suppose $b<\infty$, here the result is a quite obvious because the interval $[a,b]$ would be compact in $\mathbb{R}$ and therefore both integrals are equal for all $x\in[a,b]$ thus for limit also works. But in the other case $b=\infty$, here the integral is improper and there are cases where Riemann improper integral exist but the Lebesgue integral doesn't. Any hint is welcome.
Best Answer
Assuming $f$ is Riemann integrable on any finite interval $[a,x]$ where $x < b \leqslant +\infty$, the Riemann and Lebesgue integrals coincide, with
$$\int_a^x f(t) \, dt = \int_{[a,x]} f \, d\mu= \int_{[a,b]}f \chi_{[a,x]} \, d\mu$$
Since it is given that $f$ is nonnegative we can apply the monotone convergence theorem to obtain
$$\lim_{x \to b-}\int_a^x f(t) \, dt = \lim_{x \to b-}\int_{[a,b]}f \chi_{[a,x]}\, d\mu = \int_{[a,b]}f \, d\mu$$