If $f\in\mathcal{M}$ then $f=\sum_{n=0}^{\infty}a_n \mathcal{X}_{A_{n}}$

Prove that every measurable function $f:E\rightarrow [0,+\infty] , \:E\in\mathcal{M}$ can be written as $f=\sum_{n=0}^{\infty}a_n \mathcal{X}_{A_{n}}$ where $0\leq a_n<\infty $ and $A_n\in\mathcal{M}$

Because $f$ is measurable, we know it exists a positive increasing sequence of simple function $\{\phi_n\}$ such that $ \phi_n\ \rightarrow f $.

Consider $\psi_n=\phi_n \mathcal{X}_{[-n,n]}$, thus if we let $a_n=\phi_n$ and $A_n=[-n,n]$

We get that $f=\sum_{n=0}^{\infty}a_n \mathcal{X}_{A_{n}}$.

Is this correct ?

$a_{n}$ is required to be a number, not a function, so you cannot simply let the function $\phi_{n}$ to be $a_{n}$.

Since you let $(\phi_{n})$ to be an increasing of simple functions such that $\phi_{n}(x)\uparrow f(x)$ pointwise, the function $\phi_{n}-\phi_{n-1}$ is still a simple function by letting $\phi_{0}=0$.

Since $\phi_{n+1}\geq\phi_{n}$ for $n\geq 0$, the nonnegative simple function $\phi_{n+1}-\phi_{n}$ can be written as a form that $\sum_{k=0}^{N_{n}}\lambda_{k,n}\chi_{A_{k,n}}$ with $0\leq\lambda_{k,n}<\infty$.

Then \begin{align*} f(x)=\sum_{n=0}^{\infty}(\phi_{n+1}(x)-\phi_{n}(x))=\sum_{n=0}^{\infty}\sum_{k=0}^{N_{n}}\lambda_{k,n}\chi_{A_{k,n}}(x). \end{align*} Now we can freely enumerate $\{\lambda_{k,n}\}_{0\leq k\leq N_{n},n\geq 0}$ and $\{\chi_{A_{k,n}}\}_{0\leq k\leq N_{n},n\geq 0}$ since the summands in the infinite sum are all nonnegative.