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Given that $F$ is a left continuous and also montonic function $\mathbb R \rightarrow \mathbb R$ and $\Lambda $ is the Lebesgue-Stieltjes measure which is induced by $F$, how can we calculate the measures of the following sets:

$\Lambda(\{x\}),\Lambda([a,b])$ and $\Lambda((a,b)) $?

So, $$\Lambda(A) = \inf\left\{\sum_{k=1}^{\infty} \lambda([a_k,b_k));\,A \subset \bigcup \limits_{k=1}^{\infty}[a_k,b_k)\right\}$$

where $\lambda([a,b))= F(b) - F(a)$ if $b \geq a$ and $0$ else.

So if I understand it correctly, if I take a point as this set $A$, I just have to find the infimum of a cover (here the union of semi-open intervals) for this point and then just calculate $F(b) - F(a)$ So for $\Lambda(\{x\})$ couldn't I just use as cover such a thing as $$\bigcup \limits_{k=1}^{\infty} [x,2^{-k}\cdot\varepsilon]?$$

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  • $\begingroup$ According to my lecture notes it is left continuous. I have elaborated a bit more on the problem now, sorry. $\endgroup$
    – user62487
    Mar 14, 2013 at 9:09
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    $\begingroup$ You can use left continuous and right continuous functions to define the Lebesgue-Stieltjes measure. However, the intervals must be chosen accordingly, $[\cdot,\cdot)$ for left continuous and $(\cdot,\cdot]$ for right continuous $F$. $\endgroup$ Mar 14, 2013 at 10:04

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First, your cover should consist of half-open intervals, not closed intervals. So what you mean is rather: $$\bigcup \limits_{k=1}^{\infty} [x,2^{-k}\cdot\varepsilon).$$

Then you may get a problem if $F$ is discontinuous at $x$. Remember it's only left continuous, so you could have $\delta>0$ such that $F(x+\epsilon)>F(x)+\delta$ for any $\epsilon>0$. Then $$\sum_{k=1}^{\infty} \lambda([x,2^{-k}\cdot\varepsilon))$$ would not converge.

Why don't you just pick $[x,x+\epsilon)$ for $k=1$ and $[x,x)$ for $k>1$. Then you don't get into that trouble.

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