A set function $c$ defined on all of $\mathbb R$, is defined as follows. Define $c(E)$ to be infinity if $E$ has infinitely many members and $c(E)$ to be equal to the number of elements in $E$ if $E$ is finite; define $c(\varnothing)=0$. Show that $c$ is a countably additive and translation invariant set function. This set function is called the counting measure.
To prove this question, I must prove:
-
Countably Additive for two cases: infinite and finite.
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Translation Invariant for two cases: infinite and finite.
Let $E$ is infinite set.
Countably Additive: We must prove:
\begin{align}
c\left( \bigcup\limits_{n=1}^\infty E_n\right)=\sum\limits_{n=1}^\infty E_n
\end{align}
Now I try to prove
\begin{align}
c\left( \bigcup\limits_{n=1}^\infty E_n\right) = c(E_1\cup E_2\cup \ldots)=\infty=\infty+\infty+\ldots=c(E_1)+c(E_2)+\ldots = \sum\limits_{n=1}^\infty E_n
\end{align}
Translation Invariant, we must prove:
If $E\in \mathcal{P}(\mathbb R)$ and $E+x=\{x+y\mid y\in E\}$ then $c(E)=c(E+x)$.
Since $E$ is infinity then $c(E)=\infty$. Since $E$ is infinity, the set $E+x$ for all $x\in \mathbb R$ is also infinite set. Then, $c(E+x)=\infty$. So, $c(E)=c(E+x)$.
For the same way, finite case is similar.
I'm not sure with my effort. Is the prove above is true?
Best Answer
Your proof of countable additivity is wrong. You said 'Let $E$ is infinite set' and proved something that does not involve $E$.
Suppose $E_i$ are disjoint and $\bigcup_n E_i$ is a finite set . Then each $E_i$ is a finite set and, in fact, $E_i$ is empty for $i$ sufficiently large. Now $c(\bigcup_n E_i) =\sum_i c(E_i)$ is clear.
If $\bigcup_n E_i$ is an infinite set we have to show that $\sum_i c(E_i)=\infty$. Prove this by contradiction. Suppose $\sum_i c(E_i)<\infty$. Then $c(E_i) \to 0$ which implies that $E_i$ is empty for $i$ sufficiently large and each $E_i$ is a finite set. Do you see the contradiction now?