Given a nonnegative function $f$, suppose the Riemann integral $$\int_{a+\epsilon}^{b} f(x) \ \ dx$$
exists for every $\epsilon > 0$ and approaches a finite limit as $\epsilon \rightarrow 0^+$, so that the improper Riemann integral $$\int_{a}^{b} f(x) \ \ dx = \lim_{\epsilon \rightarrow 0^+} \int_{a+\epsilon}^{b} f(x) \ \ dx$$ exists.
How can I prove that $f$ is Lebesgue integrable on $[a, b]$ and $$\int_{[a,b]} f(x) \ \ dx = \int_{a}^{b} f(x) \ \ dx$$
I don't know how to prove the Lebesgue integrability of this case. Someone could help me please?
For example I understand that if $f$ is of variable sign and if $$ \lim_{\epsilon \rightarrow 0+} \int_{a+\epsilon}^{b} |f(x)| \ \ dx = \infty $$
Then the Lebesgue integral of $f$ over $[a, b]$ fails to exist, even if the improper Riemann integral $$\int_{a}^{b} f(x) \ \ dx = \lim_{\epsilon \rightarrow 0^+} \int_{a+\epsilon}^{b} f(x) \ \ dx$$ exists.
In fact, by the existence of either of the integrals $$ \int_A f(x) \ \ d\mu \ \, \, \int_A |f(x)| \ \ d\mu $$ implies the existence of the other and summability of $f$ would imply that of $|f|$.
Thanks for your help and time.
In the nonnegative case, you can set up an approximation scheme. Define $f_n(x)$ to agree with $f$ on $[a+1/n,b]$ and be zero elsewhere. Then $f_n$ is nonnegative and converges monotonically to $f$, so that the monotone convergence theorem gives the desired result.
In the variable sign case you can still do the same construction. It will work iff $\int_a^b |f(x)| dx < \infty$ in which case the dominated convergence theorem can fill the role that the monotone convergence theorem filled in the argument above.