Let $f: \Omega\longrightarrow \mathbb{R}$ be a Lebesgue integrable function. Does $$ s_n=\sum_{-\infty}^\infty\frac{k}{2^n}\lambda\left\{\frac{k}{2^n}<f\leq \frac{k+1}{2^n}\right\} $$ converges absolutely and $$ \lim_{n\to\infty}s_n=\int_\Omega f\,d\lambda? $$
For the case $\Omega=[0,1]$, I have proved that the above is true. For example, suppose $f$ is non-negative. Then
$$ \int f\,d\lambda\geq s_n, $$ $$ \int f\,d\lambda-s_n=\sum_{k=0}^\infty\int_{\frac{k}{2^n}<f\leq\frac{k+1}{2^n}}\left(f-\frac{k}{2^n}\right)\,d\lambda\leq \frac{1}{2^n}\lambda([0,1]). $$ Hence the above goes to zero when $n$ goes to infinity.
How about the case $\Omega=\mathbb{R}$? I meet some difficulties in this situation. Thanks.