prove that $\lim_{m \rightarrow \infty} \Sigma_{k=-m^2}^{m^2}|\int^{(k+1)/m}_{k/m}f(x)dx|=\|f\|_{L^1 (\Bbb R)}$.

Question

Suppose $f \in L^1 (\Bbb R)$, prove that $$\lim_{m \rightarrow \infty} \sum_{k=-m^2}^{m^2}\left|\int^{(k+1)/m}_{k/m}f(x)\,dx\right|=\|f\|_{L^1 (\Bbb R)}.$$

For this one, it's easy to prove when $f$ is nonnegative. What about the general case?

Answer 1

First show it's true if, say, $f$ is continuous with compact support.

Now define $$N_m(f)=\Sigma_{k=-m^2}^{m^2}\left|\int^{(k+1)/m}_{k/m}f(x)dx\right|.$$

Note that $$N_m(f)\le||f||_1.$$

Suppose $f\in L^1$ and $\epsilon>0$. Let $g$ be continuous with compact support such that $||f-g||_1<\epsilon$. We have $N_m(g)\to||g||_1$, $|N_m(f-g)|<\epsilon$ and $\left|\,||f||_1-||g||_1\right|<\epsilon$. Hence $\limsup|N_m(f)-||f||_1|\le\epsilon$.