Basic Queries related to indicator function (explanation needed) I have an indicator function for a random variable $X$ defined as follows $$1_{X\leq c}= 1 \quad \text{if }X\leq c \text{ and } 0 \text{ otherwise}$$ Now I have following queries about it.
1- Is it true that $$E[1_{X\leq c}]=1-F_{X}(c)$$ where $F_{x}(x)$ is the CDF of $X$.
2- I am interested in finding $p_X(x|X\leq c)$ which can be obtained using following steps $$p_X(x|X\leq c)=\lim_{h\to 0}\frac{P(X\in ([x,x+h)|X\in (-\infty,c])}{h}$$ $$=\lim_{h\to 0}\frac{P(X\in[x,x+h),X\in (-\infty,c])}{hP(X\in (-\infty,c])}$$ $$=\lim_{h\to 0}\frac{P(X\in [x,
\min(x+h,c)),X\in (-\infty,c])}{hP(X\in (-\infty,c])}$$ $$=\frac{p_X(x)1_{X\leq c}}{P(X\in (-\infty,c])}$$ In this I do not understand how indicator function appears in the last equation. I have following explanation for each step above. First step is a type of differentiation of Probability at value $x$. Second step is the application of Bayes rule. Third step make sure that $X$ does not have a value greater than $c$. But I do not understand the last step. Please provide some explanation. Thank you.
3- Suppose $X$ and $Y$ are some independent random variables and $A$ is some event that $XY<t$ where $t$ is some constant. Then how the following equality is true $$E[Y|X,1_A]=E[Y|X,A]$$ where $E$ denotes the expectation operator. This actually means that $p_Y(y|X,1_{A})=p_Y(y|X,A)$. So I do not understand how $p_Y(y|X,1_{A})=p_Y(y|X,A)$. I will be very thankful to you for your explanation.
 A: 1) No. $E(1_{X\le c}) = P(X\le c) = F_X(c).$ In general for an event $A,$ $E(1_A) = P(A).$
2) This last line is wrong. $1_{X\le c}$ is a random variable whereas what you're computing is a function of $x$ and $c,$ and not a random quantity. It should be $1_{x\le c}$ with a lowercase $x,$ which is just a step function (when viewed as a function of $x$ with a parameter $c.$) 
I would not refer to the second line as an application of Bayes' rule, but rather as an application of the definition of conditional probability, although the line between these is somewhat fuzzy. 
For an explanation of the corrected last step, observe that if $x\ge c,$ then $[x,\min(x+h,c))$ is the empty set and $P(X\in \emptyset)$ is zero. Whereas, if $x<c$ then for some sufficiently small $h,$ we have$x+h<c.$ Then $\min(x+h,c) = x+h$ and also $X\in [x,x+h)$ implies that $X\le c$ this event is redundant and we have $\frac{1}{h}P(X\in[x,x+h),X\le c) = \frac{1}{h}P(X\in[x,x+h))\to f_X(x).$
3) $E(Y\mid X,1_A) = E(Y\mid X,A)$ is not true. They are not even the same type of object. One is conditioned on two random variables and the other is conditioned on a random variable and an event. Perhaps you meant $E(Y\mid X,1_A=1) = E(Y\mid X,A).$ This is true because $1_A=1$ and $A$ are the same event. 
And the equality of conditional densities does not follow from the equality of conditional expectations. This is like saying just cause two things have the same average, they have the same distribution... definitely not true. Both would hold here (with the above modification that you replace $1_A$ with $1_A=1$) but that's only cause the two events you are conditioning $Y$ are literally the same thing, so both equalities hold tautologically.
