# Proof that function $\mu(y,A)$ is a probability measure for a.e. $y$

This question is related to excercise 5.1.13 from Durrett's Probability Theory and Examples.

Let $X$ and $Y$ be random variables with joint density $f(x,y)>0$. I want to show that for a.e. $y$ the function $$\mu(y,A)=\int_{A} f(x,y)\,dx \bigg/ \int f(x,y) \, dx$$ defines a probability measure for every $A$ in the borel sets $\mathcal{B}$. To do this I need to verify for a.e. $y$:

1. $\mu(y,A)\geq \mu(y,\emptyset)=0$ for every $A\in \mathcal{B}$

2. If $A_{i}\in \mathcal{B}$ is a countable sequence of disjoint sets, then $$\mu \left(y,\bigcup_{i} A_{i} \right) = \sum_{i} \mu(y,A_{i})$$

3. $\mu(\mathbb{R})=1$

I think I am able to verify these conditions, but I am worried about the "a.e. $y$" part. In particular, I have no intuition as to what the exceptional sets will be for this problem.

(For example, is it possible there exists $y$ such that condition 1 does not hold? Condition 2? Condition 3? Do I need to "keep track" of the exceptional sets to ensure that this really holds a.e.? I dont know!)

If someone can provide a rigorous proof that $\mu$ defines a probability measure for a.e. $y$, I would be much obliged.

• Do you have $f(x,y)>0$ pointwise or only for a.e. $(x,y)$ in the product space? – grndl Mar 19 '17 at 14:23
• @aduh Unfortunately the exercise does not say which of those is meant – möbius Mar 19 '17 at 14:25