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Suppose that ${\cal F}$ is a $\sigma$-algebra on a set $X$ and $\mu\mathop:{\cal F} \to [0,\infty]$ satisfies the conditions:

  1. $\mu(\emptyset) = 0$.
  2. For every pair $A$ and $B$ of disjoint sets in ${\cal F}$, $\mu(A \cup B) = \mu(A) + \mu(B)$.
  3. For every decreasing sequence $\{E_n\}$ in ${\cal F}$ (that is $E_{n+1} \subseteq E_n$ for all $n$) such that ${\bigcap_{n =1}^{\infty} E_n = \emptyset}$, we have $\lim_{n \to \infty} \mu(E_n) = 0$.

Prove that $\mu$ is a measure on ${\cal F}$.

Here's my attempt:

Proof.

Let $\{E_n\}$ be a countably infinite collection of sets such that $E_i \cap E_j =\emptyset$ for all $i,j$. Write $$E = \bigcup_{n=1}^\infty{E_n}$$ and let $$F_n = E \setminus\bigcup_{k=1}^n{E_k}.$$ for $n\geq 1.$ Then we have $$ F_{n+1}= E \setminus\bigcup_{k=1}^{n+1}{E_k} \subseteq E \setminus\bigcup_{k=1}^n{E_k}=F_n $$ and $$\bigcap_{n=1}^\infty{F_n} = \emptyset.$$ Hence, by applying condition (2), we have \begin{align*}\mu\left(F_n\right) & =\mu\left(E \setminus\bigcup_{k=1}^n{E_k}\right)\\ & = \mu(E) - \mu\left(\bigcup_{k=1}^n{E_k}\right)\\ & = \mu(E) - \sum_{k=1}^n{E_k} \end{align*} and the above holds for all $n\in \mathbb{N}$. Thus, applying condition (3), we have \begin{align*}\mu(E) & = \lim_{n\to \infty}\mu(F_n) + \lim_{n\to \infty}\sum_{k=1}^n{E_k}\\ & = \sum_{k=1}^\infty{E_k}. \end{align*} This shows that $\mu$ is a measure.

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  • $\begingroup$ I haven't made any progress. I know that it is sufficient to show that for any countably infinite collection $\{E_n\}$ of pairwise disjoint sets in $\cal{F}$, we have $\mu(\bigcup_{n=1}^\infty{E_n}) = \sum_{n=1}^\infty{\mu(E_n})$. I am not sure how the other hypotheses fit into this. I think I need to write $\bigcup_{n=1}^\infty{E_n}$ as a sequence satisfying property (3). $\endgroup$ – johnny133253 Feb 6 '19 at 6:53
  • $\begingroup$ In order to apply the conditions, you need to build a decreasing sequence of sets. How can you build a decreasing sequence of sets based on the sets $E_n, \bigcup_{n} E_n$? $\endgroup$ – supinf Feb 6 '19 at 6:57
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Your proof is correct.

There are some minor typos in some places, where you wrote $E_k$ instead of $\mu(E_k)$, but I am sure you meant the right thing.

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