A necessary and sufficient condition for a measure to be continuous. If $(X,\mathcal{M})$ is a measurable space such that $\{x\}\in\mathcal{M}$ for all $x\in$$X$, a finite measure $\mu$ is called continuous if $\mu(\{x\})=0$ for all $x\in$$X$.
Now let $X=[0,\infty]$, $\mathcal{M}$ be the collection of the Lebesgue measurable subsets of $X$. Show that $\mu$ is continuous if and only if the function $x\to\mu([0,x])$ is continuous.
One direction is easy: if the function is continuous, I can get that $\mu$ is continuous. But the other direction confuses me. I want to show the function is continuous, so I need to show for any $\epsilon>0$, there is a $\delta>0$ such that $|\mu([x,y])|<\epsilon$ whenever $|x-y|<\delta$.But I can't figure out how to apply the condition that $\mu$ is continuous to get this conclusion.
 A: To show that $f$ is continuous at $x \in [0,\infty]$, it suffices to see that the left/right limits exist and are equal to the value of the function. Take any sequence $x_n \searrow x$. Then 
$$
\mu([0,x_n]) = \mu([0,x]) + \mu([0,x_n]) - \mu([0,x]) = \mu([0,x]) + \mu(]x,x_n]) \longrightarrow \mu([0,x]) + \mu(\varnothing) = \mu([0,x])
$$
because measures have the following property (which is usually called "continuity"... let's not use that word for now) : when $\mu(A_1) < \infty$ and $A_1 \supset A_2 \supset \dots \supset A_n \supset \dots, $ then
$$
\lim_{n \to \infty} \mu(A_n) = \mu \left( \bigcap_{n=1}^{\infty} A_n \right).
$$
In our case, $]x, x_n]$ is a decreasing sequence of subsets because $x_n \searrow x$, so the property applies. Note that I need the fact that $\mu$ is a finite measure to do so, because I need to ensure $\mu(]x,x_n]) < \infty$. 
The case where $x_n \nearrow x$ is treated similarly. 
Hope that helps,
A: It seems like the contrapositive is a good way to go.  Suppose that $x\mapsto\mu([0,x])$ is not continuous, say at the point $x_0$.  Then there exists an $\epsilon>0$ such that for all $\delta>0$ there is a $y$ such that $\vert x_0-y\vert<\delta$ but $\vert\mu([x_0,y])\vert\geq\epsilon$.  Thus we can construct a sequence $(y_n)$ which converges to $x_0$, but such that $\vert\mu([x_0,y_n])\vert\geq\epsilon$ for all $n$.  Hence we can conclude that $\mu(x_0)\geq\epsilon>0$.  Hopefully you can fill in the details from there!
