# Prove $\sum_{i=1}^{\infty}P[A{\cap}C{\cap}Bi] = P[A{\cap}C]$.

How to prove this mathematically rather than using venn diagram? $$\sum_{i=1}^{\infty}P[A{\cap}C{\cap}Bi] = P[A{\cap}C]$$ where $B_i$ is the partition of $C$ (not $S$). $A$ and $C$ are arbitrary sets.

I have proved that $$A{\cap}C = A{\cap}C{\cap}C = A{\cap}C{\cap}\left({\bigcup_{i=1}^{\infty}}B_i\right) = {\bigcup_{i=1}^{\infty}}(A{\cap}C{\cap}B_i).$$ Thus $$P(A{\cap}C) = P\left({\bigcup_{i=1}^{\infty}}(A{\cap}C{\cap}B_i)\right)$$ Then I need to apply the axiom of probability/measure to get the summation out. However how do I know each $A{\cap}C{\cap}B_i$ are disjoint with each other using a formal proof instead of Venn diagram?

• Try to adopt the formatting I used. It's less cumbersome for you, and easier to read overall.
– Em.
Commented Mar 9, 2016 at 22:45

If you really want to want to show that are disjoint, then I think you can say that an arbitrary element in $ACB_i$ is by definition in $B_i$, and an arbitrary element in $ACB_j$ is in $B_j$ by definition, for $i\neq j$. These elements cannot be the same since $B_j$ and $B_i$ are disjoint. So $ACB_i$ and $ACB_j$ cannot share an element, and hence are disjoint.