Suppose $\mu(E)<\infty$ and $f$ is Lebesgue integrable on $E$. Prove the Lebesgue integral $$\int f~d\mu= \lim_{m(T)\to 0}\displaystyle\sum_k \tau_k~\mu\{t_k\le f \le t_{k+1}\}.$$ $T=\{...t_{-1}<t_0<t_1<...\}$ is a countable partition of $\mathbb{R}$, $m(T)=\displaystyle\sup_k|t_{k+1}-t_k|$ and $\tau_k\in[t_k,t_{k+1})$ are aribitrary points.
So, we need to show $$\sup\{\int g~d\mu, g\le f\ \text{and}~g~\text{is simple}\}=\lim_{m(T)\to 0}\displaystyle\sum_k \tau_k~\mu\{t_k\le f \le t_{k+1}\}$$
When the partition is finite nad $f$ is bounded, $\displaystyle\sum_k \tau_k~\mu\{t_k\le f \le t_{k+1}\}$ gives us a simple funtion and I think we can use Lebesgue monotone convergence theorem. However, $f$ are not necessarily bounded and the partition is infinite. How to deal with this case?
Hint: Since $f$ is integrable, we know that $\int_E |f|\,d\mu < \infty$. Using monotone convergence, this implies that $$\int_E |f| \mathbf{1}_{|f| > N} \,d\mu \xrightarrow{N \to \infty} 0.$$ Thus, for any $\epsilon > 0$, we can find an $N$ so that $\int_E |f| \mathbf{1}_{|f| > N} \,d\mu < \epsilon$. Use this kind of argument to restrict to the case of finite partitions and $f$ bounded.
Edit: It's a bit trickier than just what my hint said, so the full solution I've put below: For convenience, assume $f$ is non-negative (by definition of Lebesgue integral, it's easy to bridge the gap between this and the general case).
Fix $\epsilon > 0$ and choose $N$ so that $\int_E f \mathbf{1}_{f > N} \,d\mu < \epsilon$. Then break up the sum: $$\sum_k \tau_k \mu\{t_k \leq f \leq t_{k+1}\} = \sum_{k:t_{k+1}\leq N} \tau_k \mu\{t_k \leq f \leq t_{k+1}\} + \sum_{k:t_{k+1}> N} \tau_k \mu\{t_k \leq f \leq t_{k+1}\}.$$
Note that the latter term satisfies $$\sum_{k : t_{k+1} > N}\tau_k \mu\{t_k \leq f\leq t_{k+1}\} = \sum_{k: t_{k+1} > N} t_k \mu\{t_k \leq f \leq t_{k+1}\} + \sum_{k: t_{k+1} > N} (\tau_k - t_{k+1}) \mu\{t_k \leq f \leq t_{k+1}\}.$$
The latter term can be bounded by $m(T)\mu(E)$ which goes to zero as $m(T) \to 0$. Note then that the first sum is less than $f \mathbf{1}_{f \geq N}$, so this shows that for any $\epsilon'>0$, we eventually have (for sufficiently small $m(T)$ that $$\sum_{k : t_{k+1} > N}\tau_k \mu\{t_k \leq f\leq t_{k+1}\} + \epsilon' \leq \int_{E} f \mathbf{1}_{f \geq N}\,d\mu.$$
Doing the same for the upper bound of $t_{k+1}$ gives that $$\sum_{k : t_{k+1} > N}\tau_k \mu\{t_k \leq f\leq t_{k+1}\} \xrightarrow{m(T) \to 0} \int_{E} f\mathbf{1}_{f \geq N}\,d\mu < \epsilon.$$
You may then use Lebesgue dominated convergence to show that $$\sum_{k :t_{k+1} \leq N} \tau_k \mu\{t_k \leq f \leq t_{k+1}\} \xrightarrow{m(T) \to 0} \int_{E} f \mathbf{1}_{f \leq N} \,d\mu$$ where the dominating function is $f+1$ for when $m(T) < 1$. Taking $\epsilon \to 0$ completes the proof.