Limit of measures is again a measure

Given a sequence $(\mu_n)_{n\in \mathbb N}$ of finite measures on the measurable space $(\Omega, \mathcal A)$ such that for every $A \in \mathcal A$ the limit
$$\mu(A) = \lim_{n\to \infty} \mu_n(A)$$
exists. I want to show that $\mu$ is a measure on $\mathcal A$.

What I managed to figure out:

  • $\mu$ is monotone, additive and – if $\lim_n \mu_n(\Omega)$ is supposed to be taken in $\mathbb R$ – then $\mu$ is also finite (I’ll assume this). Also $\mu(\varnothing) = 0$.
  • So we can wlog assume that $\mu(\Omega) = 1$ and that $\mu_n(\Omega) \le 2$ for all $n$.

Now what remains to be shown is that $\mu$ is $\sigma$-additive, or equivalently that for $A_n\downarrow \varnothing$, we have $\mu(A_n) \to 0$ (since $\mu$ is finite).

All attempts at proving this have been futile so far. I don’t seem to see the right approach.

If possible, I would like to only receive a hint rather than a full answer. But of course, I’d be quite happy with a full answer, too; if a good hint is hard to find!

Thanks a lot in advance for your help! =)

Solutions Collecting From Web of "Limit of measures is again a measure"

I think I have figured out a (completely elementary) proof for this now. I’ll prove it by contradiction:

Let $A_n$ be a sequence of pairwise disjoint sets in $\mathcal A$. Note that $\sum_{n=1}^m \mu(A_n) = \mu(\bigcup_{n=1}^m A_n) \le \mu(\bigcup_{n=1}^\infty A_n)$ for all $m\in \mathbb N$ implies
$$\sum_{n=1}^\infty \mu(A_n) \le\mu\left(\bigcup_{n=1}^\infty A_n\right)$$

To reach a contradiction suppose that this inequality was strict. We will use these $A_n$ to construct a set $B$ for which $\mu_n(B) \not \to \mu(B)$.

By our assumption on the inequality being strict, the following must hold

  • There exists $\epsilon_0>0$ such that for each $n_0\in \mathbb N$ there exist infinitely many $m\in \mathbb N$ with $$\sum_{n=n_0}^\infty \mu_m(A_n) \ge 2\epsilon_0 $$

(if this weren’t true, then for every $\epsilon > 0$ there would exist $n_0, m_0 \in \mathbb N$ such that $\sum_{n = n_0}^\infty \mu_m(A_n) < \epsilon$ for all $m>m_0$. But then, for $m>m_0$ sufficiently large, we would have
\begin{align}
\sum_{n = 1}^{\infty} \mu(A_n) &\ge \sum_{n=1}^{n_0-1} \mu(A_n)
= \lim_{m’\to\infty} \sum_{n = 1}^{n_0-1} \mu_{m’}(A_n) \\
&\ge \sum_{n=1}^{n_0-1} \mu_m(A_n) – \epsilon \ge \sum_{n=1}^\infty \mu_m(A_n) – 2\epsilon \\
&= \mu_m\left(\bigcup_{n=1}^\infty A_n\right) – 2\epsilon \overset{m\to \infty}{\longrightarrow} \mu\left(\bigcup_{n=1}^\infty A_n\right) – 2\epsilon
\end{align}
i.e. $\sum_n \mu(A_n) \ge \mu(\bigcup_n A_n)$
which we assumed was not the case.)

Using the bulleted property above, construct two increasing sequences $m_k$, $N_k$ recursively as follows: Choose $m_1$ such that $\sum_{n=1}^\infty \mu_{m_1}(A_n) \ge 2\epsilon_0$ and then choose $N_1$ sufficiently big that $\sum_{n=N_1}^\infty \mu_{m_1}(A_n) \le \epsilon_0/5$. Note that in particular: $\sum_{n=1}^{N_1} \mu_{m_1}(A_n)\ge \epsilon_0$.

Having constructed $m_1 < m_2 <\dots < m_k, \, N_1 < N_2 < \dots < N_k$, choose $m_{k+1}$ so that $$|\mu_{m_{k+1}}(A_n) – \mu(A_n)| \le \frac{2^{-(n+1)} \epsilon_0}5 \;\text{for all $n \le N_k$}, \qquad \sum_{n = N_k+1}^\infty \mu_{m_{k+1}}(A_n) \ge 2\epsilon_0$$ and then choose $N_{k+1}>N_k$ such that $$\sum_{n=N_{k+1}}^\infty \mu_{m_{k+1}}(A_n) \le \frac{\epsilon_0}{5}$$ Note again, that this implies $\sum_{n = N_k+1}^{N_{k+1}} \mu_{m_{k+1}}(A_n)\ge \epsilon_0$.

Now we are ready to construct $B$. It is defined as $$B = \bigcup_{k\in 2\mathbb Z_+} \bigcup_{n = N_k+1}^{N_{k+1}} A_n$$

Given an odd number $k\ge1$, we have

\begin{align}
|\mu_{m_{k+1}}(B) – \mu_{m_k}(B)| &= \left|\sum_{l\in 2\mathbb Z_+} \sum_{n=N_l+1}^{N_{l+1}} (\mu_{m_{k+1}}(A_n) – \mu_{m_k}(A_n))\right| \\
&\ge \left|\sum_{n=N_k+1}^{N_{k+1}} \mu_{m_k}(A_n)\right| – \left|\sum_{n = N_k+1}^{N_{k+1}} \mu_{m_k}(A_n)\right| \\
& \quad – \left|\sum_{n = N_{k+1}+1}^\infty \mu_{m_k}(A_n)\right| -\left|\sum_{n = N_{k+1}+1}^\infty \mu_{m_k}(A_n)\right| \\
&\quad – \sum_{n=1}^{N_{k-1}} \underbrace{\left|\mu_{m_{k+1}}(A_n) – \mu_{m_k}(A_n)\right|}_{\le 2^{-n}\epsilon_0/5} \\
&\ge \epsilon_0 – \frac{\epsilon_0}5 – \frac{\epsilon_0}5- \frac{\epsilon_0}5 – \frac{\epsilon_0}5\\
&= \frac{\epsilon_0}5
\end{align}

Therefore $\left(\mu_{m_k}(B)\right)_{k\in \mathbb N}$ is not a Cauchy sequence, contradicting the fact that $\mu_n(B) \to \mu(B)$.

This concludes the proof that $\mu$ must be $\sigma$-additive.