Conditional expectation given event and random variable Let $W$ be a random variable that only takes on the values $1$ or $0$. Let $X$ and $Y$ be two other random variables. I came across the following:
$$\mathbb{E}(Y|W=1, X)$$
How is this 'conditional expectation' even defined? It doesn't seem to fit the usual definition of a conditional expectation because $W=1$ is an event while $X$ is a random variable.
I was hoping to find a definition similar to that of $\mathbb{E}(Y|X)$, which has the property that $$\int_{B} \mathbb{E}(Y|X) dP=\int_{B} Y dP$$
for all $B \in\sigma(X)$.
Additional context:
Suppose we have observations  $Y_i=W_i Y_i(1)+(1-W_i)Y_i(0)$ where the random variable $W_i \in\{0,1\}$ and the two random variables $Y_i(0), Y_i(1)$ can  each be described by a regression model, that is $$Y_i(w)=\beta_{w,0}+\beta_{w,1}X_i+\epsilon_{w,i}, \hspace{5mm} w\in\{0,1\}$$
When the author writes $E(Y_i|W_i=1, X_i)$, I am pretty sure he means that $$E(Y_i|W_i=1, X_i)=\beta_{1,0}+\beta_{1,1}X_i$$
but I am wondering if this conditional expectation is well defined or made up notation.
 A: This is a fairly standard notation, $\mathbb{E}[Y|W,X]$ being the random variable derived from the pair $(W,X)$. When the realisation of $W$ is one (1) and the realisation of $X$ is $x$, this random variable has as realisation  $\mathbb{E}[Y|1,x]$, or in a clearer representation  $\mathbb{E}[Y|W=1,X=x]$. Since the only possible realisations of $W$ are $0$ and $1$, the random variable $\mathbb{E}[Y|W,X]$ can only take as values $\mathbb{E}[Y|W=0,X]$ and $\mathbb{E}[Y|W=1,X]$.
A: I will post this as an answer to my own question, but I am not entirely sure if it is correct.
$\mathbb{E}(Y|W=1, X)$ is the random variable defined by having the property:
$$\int_B \mathbb{E}(Y|W=1,X) dP=\int_B Y dP$$ for all $B$ in $\sigma(\{W=1\}\cap C: C \in\sigma(X))$. 
From this property it immediately follows that the 'conditional' tower law holds since by definition of the property above $$\mathbb{E}(I_{\{W=1\}}E(Y|W=1,X))=E(I_{\{W=1\}}Y)$$
which implies that
$$\mathbb{E}(\mathbb{E}(Y|W=1,X)|W=1)=\frac{\mathbb{E}(I_{\{W=1\}}E(Y|W=1,X))}{P(W=1)}\\=\frac{\mathbb{E}(I_{\{W=1\}}Y)}{P(W=1)}=E(Y|W=1)$$
