Consider the measurable space $(\Omega,\mathcal{A},\mu)$. Let $f$ be a measurable, non-negative and bounded function. Show that the $\mu$-integral of f is given by $$ \int f\, d\mu=\lim\limits_{n\to\infty}\sum_{k=1}^{n2^n}\frac{k-1}{2^n}\mu\left(\left\{\omega\in\Omega: \frac{k-1}{2^n}\leq f(\omega)<\frac{k}{2^n}\right\}\right). $$
Hey! Here's my proof. Would you please tell me, if it is okay?
Define $$ A_{k,n}:=\left\{\omega\in\Omega|\frac{k-1}{2^n}\leq f(\omega)<\frac{k}{2^n}\right\} $$ and with that $$ f_n:=\sum_{k=1}^{n2^n}\frac{k-1}{2^n}1_{A_{k,n}}. $$ $f_n$ is measurable, because $A_{k,n}\in\mathcal{A}$, because of the measurability of $f$. Moreover, $f_n$ is non-negative and only takes finite many values; so $f_n$ is an elementary function.
Furthermore, $A_{k,n}\cap A_{j,n}=\emptyset$ for $j,k\in\left\{1,\ldots,n2^n\right\}, j\neq k$ and $$ A_{k,n}=A_{2k-1,n+1}\uplus A_{2k,n+1}\text{ for }k=1,\ldots,n2^n. $$ So consider any $n$ and any $x\in A_{k,n}$ for any $k\in\left\{1,\ldots,n2^n\right\}$. Then $$ f_n(x)=\frac{k-1}{2^n} $$ and $$ f_{n+1}(x)=\frac{k-1}{2^{n-2}}\vee f_{n+1}(x)=\frac{k}{2^{n-2}}-\frac{1}{2^{n-1}}, $$ so $f_n(x)\leq f_{n+1}(x)$, i.e. $(f_n)_{n\geq 1}$ increases.
Consider $n>f(x)$ for any $x\in\Omega$ (this works, because $f$ is bounded). Then $$ f_n(x)\leq f(x)<f_n(x)+2^{-n}. $$ So for $n\to\infty$ it is $2^{-n}\to 0$ and therefore $$ f_n(x)=f(x). $$ So $f_n\uparrow f$.
Now I apply monotone convergence and the definition of the Lebesgue-integral for elementary functions, getting $$ \int f\, d\mu=\lim_{n\to\infty}\int f_n\, d\mu=\lim_{n\to\infty}\sum_{k=1}^{n2^n}\frac{k-1}{2^n}\mu(A_{k,n}). $$
$\diamondsuit$