Does absolute improper Riemann-integrability imply that Riemann-integral equals Lebesgue integral?

Question

If I know that $f\colon\mathbb{R}\to\mathbb{R}$ is Riemann-integrable on every compact interval (let's say it is continuous) and that $$ \int_{-\infty}^\infty|f(x)|dx<\infty $$ as an improper Riemann integral, does it follow that $$ \int_{\mathbb{R}}f(x)dx=\int_{-\infty}^\infty f(x)dx $$ where the left hand side is Lebesgue integral and right hand side is improper Riemann integral? My purpose would be to apply dominated convergence to differentiate under the Riemann integral sign. As a side note: does the dominated convergence hold for Riemann-integrable functions if I know that the sequence converges to a Riemann-integrable limit?

Answer 1

$\int_{-\infty}^{\infty}|f(x)|dx=\lim_{N\to\infty}\int_{-N}^N|f(x)|dx$ by the definition of Riemann integrability(improper). But the lebesgue integral agrees with Riemann integral on compact sets $[-N,N]$. Thus by monotone convergence theorem, we find Lebesgue integral $\int_\mathbb{R}|f(x)|<\infty$. Thus we see by dominated convergence theorem($f$ is dominated by $|f|$, $\int_\mathbb{R}f(x)=\lim_{N\to \infty}\int_{-\infty}^\infty f(x)dx$. Thus the result follows directly.