Let $X$ be a set, $S$ = {$\emptyset$, X} and define $\mu(\emptyset) = 0$, $\mu(X) = 1$. Determine the outer measure $\mu^*$ induced by the set function $\mu: S \rightarrow [0, \infty)$ and the $\sigma - algebra$ of measurable sets.


Def. Let $\mu : S \rightarrow [0, \infty]$. S is a collection of subets of $X$. We define $\mu^*(\emptyset)=0$ and $\mu^*(E) = inf \sum_{k=1}^\infty \mu(E_k)$ where $ E \subset \bigcup_{k=1}^\infty E_k$, $E_k \in S$. If there is no countable collection of $E_k \in S$ such that $ E \subset \bigcup_{k=1}^\infty E_k$, then we define $\mu^*(E) = + \infty$. Then, $\mu^*$ is an outer measure.

Since $S$ = {$\emptyset$, X} and $\mu(\emptyset) = 0$, $\mu(X) = 1$, then $\mu(S)=1$. (Does this require proof?)

So, take $E_k \in S$, then $\sum_{k=1}^\infty \mu(E_k) \leq \mu(S) = 1$. Since $\mu(S) < \infty$, there is a countable collection of $E_k \in S$ such that $ E \subset \bigcup_{k=1}^\infty E_k$ and, thus, $\mu^*(E) < \infty$.

Since, $\mu^*(E) = inf \sum_{k=1}^\infty \mu(E_k)$, $\mu^*(E) = inf \sum_{k=1}^\infty \mu(E_k) \leq \sum_{k=1}^\infty \mu(E_k) \leq \mu(S) = 1$.

Is this on the right track?


This doesn't look right... (In particular, $S$ is not a subset of $X$, so you can't apply $\mu$ to it.)

What you need to do is select $E\subset X$ and then compute $$ \mu^*(E)=\text{inf}\Bigl\{\ \textstyle\sum\limits_{k=1}^\infty \mu(E_k) : E\subset\bigcup_{k=1}^\infty E_k, E_k\in S\Bigr\}. $$

Towards this end, note that if $E_k\in S$, then either $E_k=\emptyset$ or $E_k=X$.

Now let $E\subset X$ be non-empty. Let's find $\mu^*(E)$. By taking $E_1=X$ and $E_n=\emptyset$ for $n>1$, we have $E\subset \bigcup_{k=1}^\infty E_k$ and $\sum\limits_{k=1}^\infty u(E_k)=1$. In fact, you can easily show that the infimum is achieved here; thus, $u^*(E)=\sum\limits_{k=1}^\infty u(E_k)=1$.

You should verify that $\mu^*(\emptyset)=0$.

So $\mu^*(E)=\cases{1, & if $E\ne\emptyset$\cr 0, & if $E=\emptyset$}$. This defines the outer measure

For the second part of the problem, recall that a set $E$ is measurable if $$\mu^*(T)= \mu^*(T\cap E)+\mu^*(T\cap E')$$ for all $T\subset X$.

In particular, if $E$ is measurable, we must have $$ 1=\mu^*(X)= \mu^*(X\cap E)+\mu^*(X\cap E'). $$ You should be able to determine the measurable sets from the above.

  • $\begingroup$ Thanks for the helpful post David. When you say "in fact, you can easily show that infimum is achieve here...", would there really be any "showing" necessary? If the summation equals 1, then the infimum of the summation still equals 1, correct? $\endgroup$ – Jake Casey Feb 13 '13 at 21:24
  • $\begingroup$ @JakeCasey Yes, but it should be at least pointed out that other summations (take two of the $E_k$'s to be $X$ and the rest empty, e.g.) would be at least one. So the infimum is indeed $1$. $\endgroup$ – David Mitra Feb 13 '13 at 21:27
  • $\begingroup$ David, how would our answer change if $X$ is an interval, like: $S$ = {$\emptyset$, [1,2]} of subsets of $R$ where $\mu(\emptyset) = 0$, $\mu([1,2]) = 1$. Wouldn't our answer be largely the same since the measure is still 1? $\endgroup$ – Jake Casey Feb 14 '13 at 19:39

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