Let $X\in L^1$ be a $\mathcal F$-measurable random variable and $\mathcal G$ be a sub σ-algebra of $\mathcal F$. We say a $\mathcal G$-measurable random variable $\mathbb E[X|\mathcal G]\in L^1$ is the conditional expectation of $X$ w.r.t. $\mathcal G$ if for all $A\in\mathcal G$ we have $\int_A\mathbb E[X|\mathcal G]{\rm d}\mathbb P=\int_A X{\rm d}\mathbb P$.
Now consider this example: I roll a toss and denote by $X$ the result. Let $\Omega=\{1,2,\dotsc,6\}$ be sample space and $\mathcal F=2^\Omega$ so $X$ is clearly $\mathcal F$-measurable. Take 3 events from $\mathcal F$, says, $A_0=\{3,6\},\,A_1=\{1,4\},\,A_2=\{2,5\}$ so that $\mathcal G=\sigma(\{A_0,A_1,A_2\})$ is a sub σ-algebra of $\mathcal F$. Denote by $\mathbb E[X|\mathcal G]$ $$\mathbb E[X|\mathcal G](\omega)=\begin{cases}\frac{3+6}{2}=4.5&\omega\in A_0\\\frac{1+4}{2}=2.5&\omega\in A_1\\\frac{2+5}{2}=3.5&\omega\in A_2\end{cases}$$ It's clear that $\mathbb E[X|\mathcal G]$ is the conditional expectation of $X$. But if we take the event $A_{0,1}=A_0\cup A_1\in\mathcal G$, can $\mathbb E[X|\mathcal G]$ tells the expectation of $X$ given $A_{0,1}$?
On the one hand, it's not difficult to get the condition expectation given $A_{0,1}$ by $$\frac{1+3+4+6}{4}=\frac72=3.5$$ On the other hand, if we integrate $\mathbb E[X|\mathcal G]$ on $A_{0,1}$ we get $$\int_{A_{0,1}}\mathbb E[X|\mathcal G]\;{\rm d}\mathbb P=\int_{A_{0,1}}X\;{\rm d}\mathbb P=\mathbb E[X\cdot\mathbb I_{A_{0,1}}]=\frac{7}{3}$$ which is not the true value $3.5$.