If $f$ is Riemann int. on $[0,b)$ for all $b>0$ and Lebesgue int. on $[0,\infty)$ then $\int_{[0,\infty)}f\,dm = \lim_{b\to\infty}\int_0^bf(x)\,dx$

Question

I'm trying to understand a relation between a certain improper Riemann integral and the corresponding Lebesgue integral.

In particular, Folland claims that if $f$ is Riemann integrable on $[0,b)$ for all $b>0$ and Lebesgue integrable on $[0,\infty)$ then $\int_{[0,\infty)}f\,dm = \lim_{b\to\infty}\int_0^bf(x)\,dx$. He claims this is a consequence of the dominated convergence theorem. I fail to see why. Using the dominated convergence theorem I have shown that indeed $\int_{[0,\infty)}f\,dm = \lim_{n\to\infty}\int_0^nf(x)\,dx$. I do not see why $\lim_{b\to\infty}\int_0^bf(x)\,dx = \lim_{n\to\infty}\int_0^nf(x)\,dx$.

The dominated convergence theorem makes an assertion about limits of sequences of integrals whereas what we have on the left hand side of the last equation is the limit of integrals defined for each real non-negative $b$.

Answer 1

$\int_{[0,\infty)} fdm=\lim_{n \to \infty} \int_0^{b_n} fdm=\lim_{n \to \infty} \int_0^{b_n} f(x)dx$ for any sequence $b_n$ tending to $\infty$. This implies that $\int_{[0,\infty)} fdm=\lim_{b \to \infty} \int_0^{b}f(x)dx$.