Exercise 1.3.3 in Klenke's Probability Theory: A Comprehensive Course is as follows:
Let ($\mu_{n})_{n=1}^{\infty}$ be a sequence of finite measures on the measurable space $(\Omega, \mathcal{A})$. Assume that for any A $\in \mathcal{A}$ there exists a limit $\mu(A):=lim_{n \rightarrow \infty} \mu_{n}(A)$. Show that $\mu$ is a measure on $(\Omega, \mathcal{A})$. Hint: In particular, one has to show that $\mu$ is $\emptyset$-continuous.
It was quite straightforward to show that $\mu(\emptyset) = 0$. I tried to use the hint, and end up with one potential ambiguity. Firstly, I am not sure how to interpret the fact that a limit exists since I am not sure if we can consider converging to $\infty$ as a limit (by that point the extended real numbers are still not introduced, they will be introduced one section later). If I assume that the limits are finite real numbers, then I proceed by contradiction as follows: Let $(A_{m}))_{m=1}^\infty$ be a sequence of sets such that $\cap A_{m} = \emptyset$. Then we need to show that $lim_{m \rightarrow \infty} \mu(A_{m})$ = $lim_{m \rightarrow \infty} lim_{n \rightarrow \infty} \mu_{n}(A_{m}) = 0$. Since $\mu (A_{m})$ is monotonically decreasing and lower bounded, then the limit with respect to m exists (we still do not know what it is). Assuming it is larger than 0 leads to $\lim_{m \rightarrow \infty} \mu_{n}(A_{m}) \neq 0$ for any $n$ which contradicts the $\emptyset$-continuity of the elements of the sequence of measures.
As such, the following questions arise:
- Is it justified that I assume that $\mu(\Omega)$ is finite?
- If the assumption of finiteness of $\mu$ is justified, why does one need $\mu_{n}$ to be finite? Isn't it enough to just assume $\mu$ be finite?
