Probability Measure for $X=Y$

Question

Let $X=Y$ be uniform random variables on $[0,1]$. I wonder if it's possible to define a probability measure rigorously on $[0,1] \times[0,1]$ so that all the Borel sets in $[0,1] \times[0,1]$ are measurable. Apparently, this problem will be trivial if image of Borel sets in $\mathbb{R}^2$ under $\phi:(x,y) \to x$ is Borel. But this is false. But is image under this projection from intersection of any Borel set with line y=x measurable? If not, what is the best way to define a measure for $(X,Y)$ that matches our intuition of dependence here? that is, discarding as few as Borel sets in our definition of measurable sets (not necessarily a sigma algebra) as possible.

To give an example, we may define measurable sets as consisting of only rectangles in $[0,1] \times [0,1]$. We know that intersection of any rectangle with $y=x is just an interval, which we can measure easily. But this restriction is quite limited. I wonder if there is a standard solution to this problem.

Answer 1

Define it on a $\pi$-system $\Pi$ of rectangles $((a,b]\times(c,d])\cap[0,1]^2\subseteq [0,1]^2$.

• Extend to a ring $\mathcal{R}$ using countable additivity, then Caratheodory's extension theorem guarantees the existence of a measure on $\sigma$-algebra
• Dynkin's $\pi$-$d$ lemma (or if you prefer to call it $\pi$-$\lambda$ lemma) to get a unique measure defined on $\sigma(\Pi)=\mathcal{B}([0,1]^2)$.

Answer 2

I think the product measure is all you need. ( I.e. for identically distributed uniform distributions)

If $X=Y$ is taken literally, only the diagonal $\Delta$ has positive measure. Then, if $\lambda$ is the Lebesgue measure on $[0,1]$, we can define $\mu(A) = \lambda(\pi[A\cap \Delta])$ where $\pi$ is one of the projections and $A$ is Borel in the square.