I have the following simple problem.

Let $g: \mathbb{R} → \mathbb{R}$ be the function $g(x)=\sum n\mathbb{1}_{[n,n+1)}(x)$ where $\mathbb{1}$ denotes the indicator function.

Compute $λ_g(A)$ for any interval $A = (a, b)$ where $λ_g(A)$ is the Lebesgue-Stieltjes outer measure on $\mathbb{R}$, defined by:

$λ_g(A)=inf\big\{\sum g(b_n)-g(a_n) : A\subset \cup (a_n,b_n) \big\}$

I have considered the cover $(a-\epsilon,b+\epsilon)$ of $A=(a,b)$ for an arbitrary $0<\epsilon<1$. Then computing the outer measure and considering the values when the sum is non-zero I get;

$g(b+\epsilon)-g(a-\epsilon)=\sum n\mathbb{1}_{[n,n+1)}(b+\epsilon) - \sum n \mathbb{1}_{[n,n+1)}(a-\epsilon)=b-a+1$

My question is, is this the right method to go about calculating this outer measure and additionally is it possible to find a closed form for $λ_g$.

Thanks in advance.

  • $\begingroup$ That is the correct method for the usual Lebesgue measure. However, you left out an $n$ in your calculation, which is what makes this different from the usual Lebesgue measure. $\endgroup$ – GEdgar Apr 7 '16 at 14:43
  • $\begingroup$ Sorry that was a typo. I have edited it accordingly $\endgroup$ – David Apr 7 '16 at 14:45
  • $\begingroup$ Now your calculation is simply incorrect. You may not assume $n=1$. In fact, the $n$ in $g(b+\epsilon)$ and the $n$ in $g(a-\epsilon)$ may be different. For a start, compute the outer measure of interval $[1/2,3/2]$. $\endgroup$ – GEdgar Apr 7 '16 at 14:54
  • $\begingroup$ Thanks for the comment. Should the last equality be: $\lfloor b \rfloor - \lfloor a \rfloor +1$ ? $\endgroup$ – David Apr 7 '16 at 15:35
  • 2
    $\begingroup$ I've spent a little while on this now and still can't get a sensible answer. Have you any more tips? Or a closed answer for me to work backwards from. $\endgroup$ – David Apr 8 '16 at 8:15

You can show that for all $n\in \mathbb{N}$: $\lambda_g(\{n\})=1$ and $\lambda_g(E)=0$ if $E\cap \mathbb{N}=\emptyset$. Using that I believe you get that $\lambda_g(E)=\#(E\cap \mathbb{N})$.


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