Let $F$ be the distribution function for the random variable $X$. Prove that $F(X = x) = F(x) - \lim_{\delta \downarrow 0} F(x-\gamma)$.
This makes intuitive sense to me because you're getting the probability that $\{X \leq x\}$ holds and then subtracting out everything underneath it with a limit. I'm just not so sure how to prove this statement. I am pretty new to measure theoretic probability, and I would really appreciate some sort of help with this question.
We have $P(X=x)=P(X\leq x)-P(X<x)=F(x)-P(X< x)$.
Now take any decreasing sequence $\delta_n\to 0$. We have $\{X<x\}=\cup_{n=1}^\infty\{X\leq x-\delta_n\}$. The sequence of sets in the union is an increasing sequence (by inclusion), and hence by basic properties of measure:
$P(X<x)=P(\cup_{n=1}^\infty\{X\leq x-\delta_n\})=\lim_{n\to\infty}P(X\leq x-\delta_n\}=\lim_{n\to\infty} F(x-\delta_n)$
Hence we get $P(X=x)=F(x)-\lim_{n\to\infty} F(x-\delta_n)$. Since this is true for any decreasing sequence $\delta_n\to 0$ the result follows.