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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.

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By the way, you mean $\{x \} \in \mathcal M$. The set $\mathcal M$ contains subsets of $X$. Same for $\mu(\{x\})$, not $\mu(x)$. And you want to show that for any $\varepsilon > 0$, there exists a $\delta > 0$ such that $|\mu(]x,y])| < \varepsilon$ whenever $0 \le y-x < \delta$ if $y - x > \delta$, otherwise you have $\mu(]y,x])$ if $-\delta < y-x \le 0$. –  Patrick Da Silva Dec 4 '12 at 2:02
@PatrickDaSilva: The braces were there, but the OP wrote {x} instead of \{x\}. I fixed it. –  Nate Eldredge Dec 4 '12 at 2:09
@Nate : Oh, then it's a good thing you did, and even more good thing that you mentioned it. But there was still an issue (an important one) with the $|\mu([x,y])| < \varepsilon$. –  Patrick Da Silva Dec 4 '12 at 2:12
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2 Answers

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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!

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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,

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Are you around? –  Pedro Tamaroff Dec 4 '12 at 2:51
@Peter : Yes, what is it? By the way, that's one thing I always wanted to do with this website ; how does one leave a message without posting on a question??? –  Patrick Da Silva Dec 4 '12 at 3:36
You can ping someone in the Mathematics room. Can you drop by? –  Pedro Tamaroff Dec 4 '12 at 3:56
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