A simple question on non- atomic measures Imagine $\mu$ is a non-atomic measure and $h=\mu([0,x])$
Why could we say the map $h$ is continuous because $\mu$ is non-atomic?
We know $\mu$ is a non-atomic measure if the measure of every singleton is zero.
And why could we say the map $h$ is monotonous on $supp$ $\mu$
$\mu$ is a borel probability invariant non atomic measure.
 A: "We know $\mu$ is a non-atomic measure if the measure of every singleton is zero.": Actually this is not so.
It's true that if $\mu$ is non-atomic then the measure of every singleton is zero, and it's not hard to show that this last condition implies $h$ is continuous. But in general the converse fails.
Say $A$ is the algebra of all $E\subset[0,1]$ such that either $E$ or $[0,1]\setminus E$ is countable; define $\mu(E)=0$ in the first case and $\mu(E)=1$ in the second case. Then $\mu$ is a measure and every singleton has measure zero, but $\mu$ is not non-atomic (in fact $[0,1]$ is an atom).
A: To get continuity, you would want to show that you for any $\epsilon > 0$ you can find $\delta > 0$ such that $\mu([x-\delta,x]) \leq \epsilon$ and $\mu([x,x+\delta]) \leq \epsilon$. This follows from the following property of measures: continuity from above.
(Continuity from above) Let $E_1, E_2, \cdots$ be a sequence of measurable sets with $E_{n+1} \subseteq E_n$. Then the intersection $\bigcap_{n=1}^{\infty} E_n$ is measurable. If moreover one of the sets $E_n$ has finite measure then $$\mu (\bigcap_{n=1}^{\infty} E_n) = \lim_{n \to \infty} \mu (E_n)$$.
Let $E_n = [x,x+\frac{1}{n}]$. The intersection of all the $E_n$ is just $[x,x]$, which has measure zero, since $\mu$ is not atomic (as David C. Ullrich noted, all we need here is that the measure of singletons is zero). I'm assuming our Borel measure is finite, so $\mu(E_n) < \infty$ for all $n$. This means that $\mu (E_n) \to 0$ as $n \to \infty$. Thus there is some sufficiently large $N$ such that $\mu ([x,x+\frac{1}{N}]) \leq \epsilon$, and hence take $\delta \leq \frac{1}{N}$. An identical argument works for $E_n = [x,x-\frac{1}{n}]$.
A: Consider $x_n \to x^+$ then
$$h(x_n)-h(x)=\mu([0,x]\cup(x,x_n])-\mu([0,x])=\mu((x,x_n])$$
we have that $(x,x_n]\downarrow \emptyset$ so $\mu((x,x_n])\to \mu(\emptyset)=0$. Consider $x_n \to x^-$, then
$$h(x_n)-h(x)=\mu([0,x_n])-\mu([0,x_n]\cup(x_n,x])=-\mu((x_n,x])$$
we have that $(x_n,x]\downarrow\{x\}$ so $\mu((x_n,x])\to\mu(\{x\})$. If the measure is non-atomic, then $\mu(\{x\})=0$ and the function $h(x)$ is continuous.
