# Thread: Showing that a specific outer measure is a measure

1. ## Showing that a specific outer measure is a measure

Problems:

a. Let X be an in infinite set and let mu be a fnite outer measure on the
subsets of X such that every set {x}; x in X is mu-measurable and there
exists a countable subset A of X with mu(X\A) = 0. Show that mu is a
measure on the sigma-algebra of all subsets of X.

b) Let mu, vu be finite outer measures on X with the property in a), that is,
there exist countable sets A, B in X with mu(X\A) = vu(X\B) = 0.
Find the Lebesgue decomposition of vu with respect to mu.

2. ## Re: Showing that a specific outer measure is a measure

Originally Posted by mgarson
Problems:

a. Let X be an in infinite set and let mu be a fnite outer measure on the
subsets of X such that every set {x}; x in X is mu-measurable and there
exists a countable subset A of X with mu(X\A) = 0. Show that mu is a
measure on the sigma-algebra of all subsets of X.

b) Let mu, vu be finite outer measures on X with the property in a), that is,
there exist countable sets A, B in X with mu(X\A) = vu(X\B) = 0.
Find the Lebesgue decomposition of vu with respect to mu.
We need to see some work friend. The first problem really is not that hard (and I don't mean that in a condescending way), have you given it a try? Can you show us something? Maybe then we'll help with the second part.

3. ## Re: Showing that a specific outer measure is a measure

Sure! Here it goes (I didn't want to post any of my stuff before because I didn't want to confuse anybody with my attempt). Well, here it is,

a. I realize that our $\mu:P(X)\rightarrow[0,\infty]$ has to satisfy the following properties:
(i) $\mu(\emptyset)=0$,
(ii) $\mu(\cup Ej)=\sum_{1}^{\infty}\mu(E_j)$ where $E_i\cap E_j=\emptyset$ if $i\neq j$.
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Now, (i) and $\mu(\cup Ej)\le\sum_{1}^{\infty}\mu(E_j)$ follow from the definition of outer measure. So all we have to show is:
$\mu(\cup Ej)\ge\sum_{1}^{\infty}\mu(E_j)$.

And then I tried to use the following but it didn't do me any good (I'm not sure if it is true):
$\mu(X\setminus A_{\mu})=0 \Rightarrow X$ is countable $\Rightarrow X=\cup_{1}^{\infty}{x_i}$.

And that's pretty much it...
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b. As for this one I really have no clue. I mean I know that I'm trying to find $\nu_{a}$ and $\nu_{s}$ such that: $\nu=\nu_{a}+\nu_{s}$ where $\nu_{a}<<\mu$ and $\nu_{s}\bot\mu$. Obviously $\nu(X\setminusA_{\mu})=\nu_{s}(X\setminus A_{\mu}))$ and $\nu_{s}(A_{\mu})=0$ since $\mu(X\setminus A_{\mu})$.