A conditional expectation problem Given $(\Omega, \mathcal{F}, \mu)$ a fixed probability space and let $\{A_i\}_{i=1}^{\infty}$  a countable, measurable partition of $\Omega$ with $\mu(A_i)>0$, all $i$. For each $A_i$ we can define 
$$P(B|A_i) := \mu_{A_i}(B) = \mu (B \cap A_i)/ \mu (A_i), \quad \forall B \in \mathcal{F}$$
and 
$$E(f | A_i ) := \int f d\mu_{A_i},\quad \forall f\in L(\Omega, \mathcal{F}, \mu)$$
I cann't from these two definitions show that:
$$E(f | A_i ) \mu(A_i) = \int_{A_i} f d\mu,\quad \forall f\in L(\Omega, \mathcal{F}, \mu)$$
How can I precede?
My attempt is 
\begin{equation}
\begin{split}
E(f | A_i ) \mu(A_i) & = \int f \, \mu(A_i) d\mu_{A_i} \\
               & =  \int f (\omega) \, \left(\int 1_{A_i}(\omega') d\mu (\omega') \right) d\mu_{A_i}(\omega) \\
 & = \,\,?
\end{split}
\end{equation}
 A: To prove such a statement, the common technique is to use so-called standard machinery:

$(1)$ Prove the statement is true for indicator functions;
$(2)$ Use linearity to prove it for all simple functions;
$(3)$ Use simple to appxomiate all the non-negative function;
$(4)$ For general function $f$, use $f=f^{+}-f^{-}$.

In almost every case, if you can prove $(1)$, then all other follows immediately.

Let us now use this technique to prove what you want.
Firstly, for any $B\in\mathcal{F}$, consider the indicator function $\mathbb{1}_{B}$. Then, we compute as follows:
\begin{align*}
\mathbb{E}(\mathbb{1}_{B}|A_{i})\mu(A_{i})=\mathbb{P}(B|A_{i})\mu(A_{i})&=\dfrac{\mu(B\cap A_{i})}{\mu(A_{i})}\mu(A_{i})\\
&=\mu(B\cap A_{i})\\
&=\int_{B\cap A_{i}}d\mu\\
&=\int_{A_{i}}\mathbb{1}_{B}d\mu.
\end{align*}
Thus, the desired identity holds for indicator function. 
Then, note that simple functions are the linear combination of indicator functions, and conditional expectation is linear, so the desired identity holds for all simple functions (you could write them out and combine them back and you will see).
Thirdly, for any non-negative function $f$, there exists an increasing sequence of non-negative simple functions that converge to $f$, namely $\{\phi_{n}\}_{n=1}^{\infty}$, $\phi_{n}\geq 0$, $\phi_{n}\leq \phi_{n+1}$ and $\phi_{n}\longrightarrow f$ as $n\longrightarrow\infty$. Then it follows from the monotone convergence theorem of conditional expectation and of regular integral that the desired results holds for all non-negative functions, since $$\mathbb{E}(f|A_{i})\mu(A_{i})=\lim_{n\rightarrow\infty}\mathbb{E}(\phi_{n}|A_{i})\mu(A_{i})=\lim_{n\rightarrow\infty}\int_{A_{i}}\phi_{n}d\mu=\int_{A_{i}}fd\mu.$$
Finally, for general integral function $f$, we can write $f=f^{+}-f^{-}$ and $f^{+}$, $f^{-}$  are non-negative. Then by the linearity of conditional expectation, it follows immediately that the desired equality holds for all integral function $f$.
