If $f \leq 0$ and and the improper Riemann integral $\int_{\infty}^{\infty}dx$ of $f$ exist then the Lebesgue integral $\int_{R} f d \mu $ always exists and equals the improper integral.
I do not understand the relation between the Riemann integration and Lebesgue integration. Please help me, if you have any good answer. I checked the Internet but I did not found any good answer.
Thank you.
On
Hints:
On any bounded interval $[a,b]$, if the Riemann integral exists then so does the Lebesgue integral and they are equal.
The improper Riemann integral would equal, say, the limit of $\int_{-n}^n f(x)\,dx$, as $n \to \infty$.
The Lebesgue integral of $f$ over $[-n,n]$ can also be viewed as the integral of $f 1_{[-n,n]}$ over $\mathbb{R}$.
The monotone convergence theorem.
For a function to be Lebesgue integrable, we need two things: it's measurable (very easy, and always true for anything we can do any sort of Riemann integral on), and it's absolutely integrable. That is, the integral of the absolute value of $f$ is finite, in whatever integration scheme we choose - because they all agree for nonnegative functions.
The improper Riemann integral is one of those schemes, taking the limit as $A,B\to\infty$ of $\int_{-A}^B f$. In our example, with $f\le 0$, $\int_{-A}^B |f|=-\int_{-A}^B f$. If the latter converges (to something finite) as $A,B\to\infty$, we have that $f$ is absolutely integrable, and that makes it Lebesgue integrable.