Show $\mu$($E$) $=$ $\sum_{x_n\in E} a_n$ is a measure on $\mathcal P$($\Omega$).

Question

Let $\Omega$ be a nonempty set, $\{x_n\}_{n}$ a sequence of distinct elements of $\Omega$, and $\{a_n\}_{n}$ a sequence of nonnegative real numbers. For E $\subset$ $\Omega$, define $\mu$($E$) $=$ $\sum_{x_n\in E} a_n$.

I'm trying to show $\mu$ is a measure on $\mathcal P$($\Omega$).

I've already shown that condition (1) holds: Since $\{a_n\}_{n}$ is a sequence of nonnegative real numbers, we have $\mu$($E$) $=$ $\sum_{x_n\in E} a_n$. $\ge$ $0$ for all $E$ $\in$ $\mathcal P$($\Omega$).

However, I'm stuck showing that conditions (2) and (3) hold.

That is, $\mu$($\emptyset$) $=$ $0$, and for a sequence $\{E_n\}_{n=1}^{\infty}$ of pairwise disjoint sets $\mu$(${\bigcup_{n=1}^{\infty}E_n}$)$=$$\sum_{n=1}^{\infty}$ $\mu$($E_n$)

Any suggestions?

Although it isn't a proof, can you show that $\mu(E\cup F) = \mu(E) + \mu(F)$ if $E$ and $F$ are disjoint? This special case can help fix ideas. — Umberto P., Mar 14 '18 at 00:43
If $E = \emptyset$ the sum defining $\mu(E)$ has no terms at all, since $x_n \notin E$ for all $n$. By convention the empty sum equals $0$. — Umberto P., Mar 14 '18 at 00:43
Possible duplicate of Proving $\sigma$-additivity and interchanging order of summation/integration just because positive — BCLC, Mar 19 '18 at 06:40

Alex Ortiz · Answer 1 · 2018-03-14T00:49:13.200

1

You have actually already asked this question here. The connection with your previous question is that we are taking $\lambda$ to be counting measure, and if $f:\Bbb \{x_n\}_n\to[0,\infty)$ is a sequence, i.e., a measurable function, with $f(x_n) = a_n$, then $\int_E f\,d\lambda = \sum_{x_n\in E} a_n$ by definition.

For (2), the empty sum is $0$ by convention. For (3), use the monotone convergence theorem.

edited Mar 14 '18 at 00:49

answered Mar 14 '18 at 00:43

Alex Ortiz

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Show $\mu$($E$) $=$ $\sum_{x_n\in E} a_n$ is a measure on $\mathcal P$($\Omega$).

1 Answers1