Conditional expectation as expectation $X$ is a real-valued random variable.
$B$ is an event.
Is it right that the conditional expectation $E[X\mid B]$ can be seen simply as an ordinary expectation over a new probability space having as a measure the corresponding conditional probability?
Thanks
 A: What @Stefan Hansen said is enough. Just wanted to add that, if $X\geq 0$ or $\mathbb{E}[|X|1_B]<\infty$, then we can write
$$ \mathbb{E}[X|B] = \mathbb{E}[X\cdot P(B)^{-1}1_B],$$
where obviously $P(B)^{-1}$ can be taken out of expectation. This makes the conditional expectation (on an event) of a variable look just like expectation of a different variable, but under the same measure. Here, $1_B$ is $1$ if $B$ takes place and $0$ otherwise.
Best regards
A: Let $(\Omega,\mathcal{F},P)$ be a probability space. Then for any event $B\in\mathcal{F}$ having positive probability we can define a new probability measure as
$$
P_{\mid B}(A):=\frac{P(A\cap B)}{P(B)},\quad A\in\mathcal{F}.
$$
It is easy to check that $(\Omega,\mathcal{F},P_{\mid B})$ indeed defines a new probability space. 
Now, let $X:\Omega\to\mathbb{R}$ be an integrable random variable, and recall that the ordinary expectation ${\rm E}[X]$ is just the integral of $X$ with respect to $P$, i.e. 
$$
{\rm E}[X]=\int_\Omega X\,\mathrm dP.
$$
Then the conditional expectation ${\rm E}[X\mid B]$ is just the integral of $X$ with respect to $P_{\mid B}$, i.e.
$$
{\rm E}[X\mid B]=\int_\Omega X\,\mathrm dP_{\mid B}.
$$
