I know that on a closed bounded interval, say $[a,b]$ in $R^1$, if a function is Riemann integral, then it is Lebesgue integrable, and the values of those two integrals are the same.
But, is this true in general??
If $A$ is some set in $\mathbb{R}^n$, and if $f$ is Riemann integrable on $A$, then is it true that $f$ is also Lebesgue integrable and the value of this integral is the same as that of Riemann integral?
It seems that it should be true when $A$ is bounded, but what happens when $A$ is unbounded, for example, if $A = \mathbb{R}^n$?
Best Answer
If $A$ is unbounded, the Lebesgue integral may not exist even though the improper Riemann integral does. For example, consider $A = [0, \infty)$ and
$$f(x) = \frac{(-1)^{\lfloor x\rfloor}}{1+\left\lfloor\frac{x}{2}\right\rfloor} = \begin{cases} \dfrac{1}{n+1} & 2n \leq x < 2n + 1\\ \dfrac{-1}{n+1} & 2n + 1 \leq x < 2n + 2. \end{cases}$$
Then $\int_0^{\infty}f(x)dx$ exists (and is equal to zero), while the Lebesgue integral doesn't exist because
$$\int_{[0,\infty)}f^+ dm = \int_{[0,\infty)}f^- dm = \sum_{n=0}^{\infty}\frac{1}{n+1} = \infty.$$