1
$\begingroup$

Can someone confirm the result of this exercise? Im not sure if I had overlook something. Thank you.

Calculate $\mu_F(\{a\})$ where $\mu_F$ is the Lebesgue-Stieltjes measure induced by a measuring-generating function $F$.

Context: here $F:\Bbb R\to\Bbb R$ is increasing and left-continuous and we set $$\nu_F([a,b)):=\begin{cases}F(b)-F(a),&b>a\\0,&\text{otherwise}\end{cases}\tag1$$

Then the Lebesgue-Stieltjes measure induced by $F$ is defined by

$$\mu_F(A):=\inf\left\{\sum_{k=1}^\infty\nu_F([a_k,b_k)):A\subset\bigcup_{k=1}^\infty[a_k,b_k)\right\}\tag2$$ for some measurable $A\subset\Bbb R$ and any countable collection of left-closed intervals $[a_k,b_k)$.

Note that for any chosen $m\in\Bbb N$ then $C_m:=\left\{\big[a-\frac1n,a-\frac1{n+1}\big):n\in\Bbb N_{\ge m}\right\}$ is a cover of $\{a\}$ because $\bigcup C_m=[a-1/m,a]$. Then we find that $$ \sum_{k=m}^\infty\nu_F\left(\Big[a-1/k,a-\frac1{k+1}\Big)\right)=\sum_{k=m}^\infty \left(F\Big(a-\frac1{n+1}\Big)-F\Big(a-\frac1n\Big)\right)\\ =\lim_{n\to\infty}F(a-1/n)-F(a-1/m)=F(a)-F(a-1/m)\tag3 $$ where in the last limit we used the left-continuity of $F$. Thus $$ \mu_F(\{a\})\le\inf\left\{\sum_{k=m}^\infty\nu_F(I_m):I_m\in C_m,\, m\in\Bbb N\right\}\\=\inf\Big\{F(a)-F(a-1/m): m\in\Bbb N\Big\}=0\tag4 $$ where we used again the left continuity of $F$, that is, for each $\epsilon>0$ there is a $N\in\Bbb N$ such that $F(a)-F(a-1/n)<\epsilon$ for all $n\ge N$. Because the Lebesgue-Stieltjes measure is non-negative then we conclude that $\mu_F(\{a\})=0$.

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

That seems awfully complicated. Just do

$$\begin{align}\mu_F(\{a\}) &= \mu_F\left(\bigcap_{n\geq 1}[a,a+1/n[\right) \\ &=\lim_{n\to+\infty}\mu_F ([a,a+1/n [) \\ &= \lim_{n\to +\infty} F(a+1/n) - F(a) \\ &= F(a^+)-F(a), \end{align} $$which need not be zero if $F$ has a jump discontinuity at $a$.

$\endgroup$
6
  • $\begingroup$ but this cannot be correct if my result is correct. Moreover: you had assume equality at first instead of inequality, how you justify that? I did this approach before but I found the other that I had written above more stronger. $\endgroup$
    – Masacroso
    May 11, 2018 at 2:16
  • $\begingroup$ "Dualize" exercise 28 in page 39 of Folland's Real Analysis book (second edition) and you get my result. I'll look at your work more carefully to see if I find any mistake. $\endgroup$
    – Ivo Terek
    May 11, 2018 at 2:23
  • $\begingroup$ Ok. $C_m$ is not a cover of $\{a\}$. This would mean that there is a certain $n\geq m$ such that $$a-\frac{1}{n}\leq a \color{red}{<} a -\frac{1}{n+1}, $$but this is impossible. It is not true that $\bigcup C_m = [a-1/m,a]$. $\endgroup$
    – Ivo Terek
    May 11, 2018 at 2:30
  • 1
    $\begingroup$ ok, I see, my mistake was write $\bigcup C_m=[a-1/m,a]$ instead of the correct one $\bigcup C_m=[a-1/m,a)$. Thank you for your time. $\endgroup$
    – Masacroso
    May 11, 2018 at 2:35
  • $\begingroup$ @IvoTerek If $\mu_F(A) > 0$ then can we say that $A$ must have open intervals then? $\endgroup$
    – user209663
    Nov 27, 2021 at 22:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .