Is my Proof Correct: $(X, \mathcal{A}, \mu)$ $\sigma-$finite measure space then $\mu^{*}(B) < \infty$

Question

My proof is different to my professor's and I wanted to ensure that mine is still correct and I am not overlooking something.

So: Let $\mu$ be a $\sigma-$finite measure on $X$. Show that for $B \in \mathcal{A^{*}}$ (which is the system of $\mu^{*}-$measurable sets), $\mu^{*}(B) < \infty$.

First the obvious: $B \subseteq X$. Moreover, from $\sigma-$finite measure we get $\exists (A_{n})_{n} \subseteq \mathcal{A}$ so that $X \subseteq \bigcup_{n}A_{n}$ and $\mu(A_{n})<\infty$.

Well $\mu^{*}(B) \leq \mu^{*}(X)$, however, from Caratheodory we know that $\mu^{*}(X)=\mu(X)$, since $ X \in \mathcal{A}$

and $\mu(X)\leq\mu(\bigcup_{n}A_{n})\leq\sum_{n}\mu(A_{n})<\infty$ q.e.d

Is this correct? Am I missing something?

Answer 1

Let's try inserting $B$ into your decomposition of $X$. Notice that if $X\subseteq \cup_{n}A_n$ as in your proof attempt, then since $B\subseteq X$, we can write $$X = B \cup \left(\bigcup_{n} (A_{n}\setminus B)\right)$$

Then $\mu (B)<\infty$ since $\mu$ is $\sigma$-finite since, and $\mu (B) = \mu^{*} (B)$ since $B \in A^{*}$ (this last claim I am unsure about)