Difference between continuity point of a measure and of a function? I was wondering what the relationship is between a continuity point of a measure and of a function? Or are they even related? A measure is of course a function in itself. Say we are taking the $\epsilon$-$\delta$-definition of continuity for a function and the definition that $x$ is a point of continuity for the measure $\mu$ if $\mu(\{x\})=0$.
If we consider  the distribution function $F(x) = \mu((-\infty,x])$, then if $x$ is a point of continuity is $\lim_{\epsilon \to 0} F(x+\epsilon) = F(x)$ or in other words that $F$ is continuous at $x$? Or is this always true?
 A: The distribution function $F(x) = \mu((-\infty,x])$ is always non-decreasing and right-continuous. On the other hand:

Let $F(x) = \mu((-\infty,x])$. If $F(x)$ is finite at $x$, then $F$ is continuous at $x$ if and only if $\mu(\{x\})=0$.

Proof
$(\Rightarrow)$ Since $F$ is is always non-decreasing, we have, for all $n$,
$$\mu ((x-1/n,x]) = F(x) - F(x-1/n) < \infty$$
Since $\{(x-1/n,x]\}_n$ is a decreasing family of set and $\bigcap_n (x-1/n,x]=\{x\}$, we have
$$ \mu(\{x\})  = \mu \left ( \bigcap_n (x-1/n,x]  \right)= \lim_n  (F(x) - F(x-1/n)) =0$$
$(\Leftarrow)$ Let $a_n$ be any sequence of positive numbers converging to $0$.
Note that $\{(-\infty,x-a_n]\}$ is an increasing family of sets and $\bigcup_n (-\infty,x-a_n] = (-\infty,x)$. So
$$ \lim_n F(x-a_n)= \lim \mu((-\infty,x-a_n])= \mu((-\infty,x))$$
Since this is true for any sequence of positive numbers converging to $0$, we have
$$\lim_{y \to x^-} F(y) = \mu((-\infty,x))$$
Now, if $\mu(\{x\})=0$, then
$$ F(x) = \mu((-\infty,x])= \mu((-\infty,x)) + \mu(\{x\}) = \lim_{y \to x^-} F(y) $$
So $F$ is left-continuous on $x$. But we know that  $F$ is right-continuous on $x$. So $F$ is continuous on $x$. $\square$
