# 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!

-
This is the Vitali–Hahn–Saks theorem. It is not that straightforward to prove. It is certainly not an easy exercise. – Michael Greinecker Mar 10 '12 at 16:26
@MichaelGreinecker: Thanks for the pointer. – Sam Mar 10 '12 at 17:20
Radon measures may clarify something! But I am also looking for a reference of that! – checkmath Mar 10 '12 at 17:38
@chessmath: I have also thought about proving it for Radon measures first. But it seems that even though one can use the Riesz representation theorem and stuff like that, it doesn't really make it easier. I guess the real reason why the limit must again be a measure lies in the fact that for $\mu_n(A)$ to converge for every $A$, the "mass distribution" of $X$ is not allowed to shift too much for different $n$. So the extra structure doesn't really help. – Sam Mar 14 '12 at 20:45

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.

-
+1, this is very clever... It is close to the hints Fremlin gives in Exercises Y246g,h,i (especially g) on page 190, volume 2. Bogachev proves the result in Theorem 4.6.3 on page 275 of volume one. Both use uniform integrability and are thus a "bit" less elementary. – t.b. Mar 16 '12 at 12:47