If $X$ is an integrable random variable and that $A$ is an event, how to show $E(X|A) = E(X|\mathbb{1}_A)$? Suppose that $X$ is an integrable random variable and that $A$ is an event. How can I show that $E(X|A) = E(X|\mathbb{1}_A)$? Is this something that is taken by definition?
Let's assume that $A$ is an event with $0&lt;P(A)&lt;1$ so that
$$
\mathrm{E}[X\mid A]:=\mathrm{E}[X\mathbf{1}_A]/P(A)
$$
as well as $\mathrm{E}[X\mid A^c]$ are well-defined. 

The identity in your post is in fact wrong but instead we have the following
$$
\mathrm{E}[X\mid \mathbf{1}_A]=\mathrm{E}[X\mid A]\mathbf{1}_A+\mathrm{E}[X\mid A^c]\mathbf{1}_{A^c}.
$$
To show this identity you have to show that the right-hand side (let's call it $Y$) satisfy the criteria for being a conditional expectation of $X$ given the random variable $\mathbf{1}_A$. That is, we need to show that $Y$ 


is $\sigma(\mathbf{1}_A)$-measurable,
is integrable,
satisfies $\int_B Y\,\mathrm dP=\int_BX\mathrm dP$ for all $B\in\sigma(\mathbf{1}_A)$.


This should be more or less straightforward once you've realized what $\sigma(\mathbf{1}_A)$ looks like.
It is not exactly true. &nbsp; However, yes, everything is basically definitional.

$\mathsf E[X\mid A]$ is the (conditional) expected value of $X$ when event $A$ occurs. &nbsp; This is a constant; a real value (assuming $X$ is a real valued random variable).

$\mathsf E[X\mid \mathbf 1_A]$ is the conditional expectation of $X$ measured against the indicator random variable of $A$. &nbsp; This is a random variable. &nbsp; That means it is a function of the support of the indicator random variable.  $$\mathsf E[X\mid \mathbf 1_A]~:~\{0, 1\}\mapsto \Bbb R$$

Thus the first expected value is actually that function evaluated with an argument of $1$. &nbsp; Also the evaluation at $0$ has similar meaning.

$$\mathsf E[X\mid A] = \mathsf E[X\mid \mathbf 1_A](1) \quad = \mathsf E[X\mid\mathbf 1_A{=}1]\\ \mathsf E[X\mid A^\complement] = \mathsf E[X\mid \mathbf 1_A](0) \quad = \mathsf E[X\mid \mathbf 1_A{=}0]~~$$