Probability and Expectation using Simple Functions I am trying to show that if I have a probability/cdf that I can rewrite it to as as an expectation of an indicator/simple function. Let $Y$ be a random variable with support on $\mathbb{R}$.  I want to show:
$$
 P(Y>a)=E[I(Y>A)] $$
By taking the expectations
$$E[I(Y>A)]= \int_{-\infty}^\infty I(Y>a)f_y(y)dy= \int_{a}^\infty f_y(y)dy$$
and using the fact that the sum of a pdf over its support is 1,
$$\int_{-\infty}^\infty f_y(y)dy=\int_{-\infty}^a f_y(y)dy +\int_{a}^\infty f_y(y)dy=1$$
We write,
$$
 P(Y>a)=1-P(Y\leq a)=1-\int_{-\infty}^a f_y(y)dy=1-\int_{\infty}^\infty f_y(y)dy+\int_{a}^\infty f_y(y)dy$$
$$=1-1+\int_{a}^\infty f_y(y)dy=\int_{\infty}^\infty I(Y>a)f_y(y)dy$$
Three questions:
1) Is my logic correct in the proof above?
2)When integrating over a simple/indicator function. What are the rules? To make it disappear do you "infuse" it into the limit of the integral?
3) Are there any other major tricks with the indicator function to make it appear/disappear? Obviously it's not differentiable so I can't take a derivative of the indicator function. 
 A: *

*Yes. I'm not sure it needs to be that complicated. The proposition seems to follow from the fact that $P(A) = E[I(A)]$ where $A = \{Y > a\}$

*Usually infuse but not necessarily right away (see 3).
Afaik, we have $P(A) = E[I(A)]$.
Generally, if we have a measure space (or probability space) $(S, \Sigma, \mu)$ and a measurable set (or event) $A \in \Sigma$, then
$$\int I(A) d\mu = \mu(A)$$
If $\mu$ is a probability measure (call it P), then
$$E[I(A)] := \int I(A) dP = P(A)$$
While probability theory considers $E[I(A)] = P(A)$ a definition, it's not difficult to prove $E[I(A)] = P(A)$:
$$E[I(A)] = 1P(A) + 0P(A^C) = P(A)$$
So instead of reinventing the wheel, I think you can just apply that fact.
A simple measurable function (or simple random variable) $f$ is a finite linear combination of indicator functions:
$$f = \sum_{i=1}^{n} a_i I(A_k)$$
where $a_i \ge 0$ (or real or complex numbers, depending on the textbook or situation) and $A_k \in \Sigma$
$$\int f d\mu = \sum_{i=1}^{n} a_i \mu(A_k)$$
or
$$E[f] := \int f d\mu = \sum_{i=1}^{n} a_i P(A_k)$$
either of which follow from the definition of integration of an indicator function.


*There are other things to do for the indicator function.


Sometimes you want to compute $E[X|A]$ where $X$ is a random variable (in $(S, \Sigma, \mu)$) and $A$ is an event ($A \in \Sigma$). By definition:
$$E[X|A] = E[X I(A)]/P(A)$$
Say for example $X = 1_B$ where $B$ is another event. Consider the term $E[X I(A)]$:
$$E[X I(A)] = \int_{\Omega} 1_B 1_A dP$$
$$= \int_{\Omega} 1_{A \cap B} dP$$
$$= P(A \cap B)$$
or (say A and B are intervals)
$$E[X I(A)] = \int_{\mathbb R} 1_B 1_A dt$$
$$= \int_{\mathbb R} 1_{B \cap A} dt$$
$$= \int_{B \cap A} 1 dt$$
$$= t|_{\inf(B \cap A)}^{\sup(B \cap A)}$$
Note that $E[X|A]$ is aka $P(B|A)$
If $B = [1,3]$ and $A = [0,2]$, then
$$E[X I(A)] = \int_{\mathbb R} 1_{[1,3]} 1_{[0,2]} dt$$
$$= \int_{\mathbb R} 1_{[1,2]} dt$$
$$= \int_{1}^{2} 1 dt$$
$$= t|_{1}^{2} = 1$$
or
$$E[X I(A)] = \int_{\mathbb R} 1_{[1,3]} 1_{[0,2]} dt$$
$$= \int_{1}^{3} 1_{[0,2]} dt$$
$$= \int_{1}^{2} 1 dt$$
$$= t|_{1}^{2} = 1$$
Consider another random variable $Y$ and another indicator function $1_C$ where $C := \{Y = 5\}$ is an event. Then
$$E[Y|C] = \int_{\Omega} Y 1_C dP$$
$$= \int_{\Omega} 5 1_C dP$$
$$= 5 \int_{\Omega} 1_C dP$$
$$= 5 P(C)$$
or (say C is an interval and Y has a pmf and has a range of \mathbb N)
$$E[Y|C] = \sum_{\mathbb N} y 1_C f_Y(y)$$
$$= \sum_{\mathbb N} 5 1_C P(Y = 5)$$
$$= 5 \sum_{5} P(Y = 5)$$
$$= 5 P(Y = 5)$$
If $D$ is an interval and event and $Z$ is a random variable that has a pdf, then
$$E[Z|D] = \int_{\mathbb R} z 1_D f_Z(z) dz$$
$$= \int_{D} z f_Z(z) dz$$
A: It's unclear what needs to be proven here, because if $B$ is an event (say $B=\left\{Y>a\right\}$), then $I(B)$ is a simple function, and $E[I(B)]$ is then defined as $P(B)$. In other words, $P(Y>a) = E[I(Y>a)]$ by definition of expectation.
OTOH, if you're being asked to prove that $P(Y>a)$ can be expressed as the expectation of $g(Y)$, where $g$ is a simple function from the reals to the reals, then
$$g(y):=I_{(a,\infty)}(y)\tag1$$
will do, because $P(Y>a)=E[I(Y>a)]$ by definition of expectation, and we have the equalities
$$I(Y>a)=I(Y\in(a,\infty))=I_{(a,\infty)}(Y)\,.\tag2$$
If also it is known that $Y$ has a density $f_Y(y)$, then you can apply the theorem
(for any function $g$)
$$
E[g(Y)] = \int_{\mathbb R} g(y)f_Y(y)\,dy\tag3$$
to express $P(Y>a)$ in terms of an integral over $\mathbb R$ of the simple function (1):
$$
P(Y>a)=E[I(Y>a)]\stackrel{(2)}=E[I_{(a,\infty)}(Y)]\stackrel{(3)}=\int_{\mathbb R}I_{(a,\infty)}(y)\,f_Y(y)\,dy
$$
To answer your question about rules for integrals that involve indicators, when you've got an indicator $I_{(a,b)}$ of a real interval, it's a matter of notational convention to write
$\int_{\mathbb R} I_{(a,b)} g(y)\,dy$
as
$\int_a^b g(y)\,dy$. So for example we typically write
$$\int_{\mathbb R} I_{(a,\infty)}(y)\,f_Y(y)\,dy
=\int_a^\infty f_Y(y)\,dy\,.
$$
