David Williams' Probability with Martingales 16.5
Q1. This is how I interpret the part before right-continuity:
$$\forall x \in \mathbb R, \exists \{y_n\}_{n=1}^{\infty}: \lim y_n = x, y_n \ge y_{n+1} \wedge \mu_X(y_n) = 0$$
Is that right?
Q2. How do I prove this please? (Ignore if interpretation is wrong I guess)
I tried: This is easy to prove for the case where $x$ is a non-atom or the set of atoms is empty or finite. For the case where $x$ is an atom and the set of atoms is countable:
Complements of countable sets are dense in $\mathbb R$. So, $$\forall \varepsilon > 0, \exists y_n \in [x,y_{n-1})$$ where $y_0=x+\varepsilon$.
ETA: Wait, how do I prove that $y_n \to x$? I guess $\inf y_n$ exists and $\lim y_n = \inf y_n$, but is $\inf y_n=x$? Suppose on the contrary that $\inf y_n > x$. Then...$[x,\inf y_n)$ has no non-atoms?
Q3. Do we have that $F(c-) = P(X<c)$ ?
Q4. Do we have that $F(c-) = \lim_{t \to c^{-}} F_X(t)$ ?
Q5. Do we have that $$\lim y_n = x, y_n \ge y_{n+1} \to y_n \ge x$$ but $$\lim y_n = x, y_n \ge x \nrightarrow y_n \ge y_{n+1}?$$ I think of $0,1,0,1/2,0,1/3,...$
Your interpretation is correct. The proof is very simple: just choose the terms of the sequence one-by-one to approach $x$ from above, avoiding any of the countably many atoms. Explicitly, start by letting $y_1$ be any non-atom which is greater than $x$. Having defined $y_n$, define $y_{n+1}$ to be some element of some element of $(x,\min(x+1/n,y_n))$ which is not an atom. Then $y_{n+1}<y_n$ so the sequence is decreasing and $x<y_{n+1}<x+1/n$ so it converges to $x$.