Folland proves the following theorem:

Let $\mathcal{A} \subset \mathcal{P}(X)$ be an algebra, $\mu_0$ a premeasure on $\mathcal{A}$, and $\mathcal{M}$ be the sigma algbera generated by $\mathcal{A}$. There exists a measure on $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$ - namely, $\mu = \mu^*|_{\mathcal{M}}$ where $\mu^*$ is the outer measure. If $\nu$ is another measure on $\mathcal{M}$ that extends $\mu_0$, then $\nu(E) \leq \mu(E)$ for all $E \in \mathcal{M}$, with equality when $\mu(E) < \infty$

I have a question about the second part of the statement, showing the equivalence of $\mu$ and $\nu$. When I did this myself, I used the following arguement:

Let $E \subset X$ and $A_j \in \mathcal{A}$, then: $$\nu(E) \leq \sum \nu(A_j) = \sum \mu_0(A_j) = \sum \mu(A_j) = \mu(\bigcup A_j) \, \,\forall j \implies \nu(E) \leq \mu(E)$$ $$\mu(E) \leq \sum \mu(A_j) = \sum \mu_0(A_j) = \sum \nu(A_j) = \nu(\bigcup A_j) \, \,\forall j \implies \mu(E) \leq \nu(E)$$

Folland does this a different way, is this correct though?


The first assertion follows from Caratheodory's theorem and proposition 1.13 since the $\sigma$-algebra of $\mu^*$-measurable sets include $\mathcal{A}$ and hence $\mathcal{M}$. As for the second assertion, if $E\in \mathcal{M}$ and $E\subset\bigcup_{1}^{\infty}A_j$ where $\{A_j\}_{1}^{\infty}\in\mathcal{A}$, then by montonocity $\nu(E)\leq \nu\left(\bigcup_{1}^{\infty}A_j\right) \leq \sum_{1}^{\infty}\nu(A_j) = \sum_{1}^{\infty}\mu_0(A_j)$, whence $\nu(E)\leq \mu(E)$. If we set $S = \bigcup_{1}^{\infty}A_j$, we have

$$\nu(A) = \lim_{n\to\infty}\nu\left(\bigcup_{1}^{n}A_j\right) = \lim_{n\to\infty}\mu\left(\bigcup_{1}^{n}A_j \right) = \mu(A)$$

If $\mu(E) < \infty$, choose the $A_j$'s such that $\mu(A)\leq \mu(E) + \epsilon$, hence $\mu(A\setminus E) < \epsilon$, and

$$\mu(E)\leq \mu(A) = \nu(A) + \nu(A\setminus E)\leq \nu(E) + \mu(A\setminus E)\leq \nu(E) + \epsilon$$

Since $\epsilon$ is arbitray, $\mu(E) = \nu(E)$. Finally, suppose $X = \bigcup_{1}^{\infty}A_j$ with $\mu_{0}(A_j) < \infty$, where we can assume that the $A_j$'s are disjoint. Then for any $E\in \mathcal{M}$, $$\mu(E) = \sum_{1}^{\infty}\mu(E\cap A_j) = \sum_{1}^{\infty}\nu(E\cap A_j) = \nu(E)$$ so $\nu = \mu$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.