I know the following theorem (see exercise 1.3.3 from Achim Klenke: »Probability Theory — A Comprehensive Course«):

Let $(\mu_n)_{n\in\mathbb{N}}$ be a sequence of finite measures on the measurable space $(\Omega,\mathcal{A})$. Assume that for any $A \in \mathcal{A}$ there exists the limit $\mu(A) := \lim_{n→\infty} \mu_n (A)$.

Then $\mu$ is a measure on $(\Omega,\mathcal{A})$.

For all $n\in\mathbb{N}$ the function $F_n(x) = \frac{n x}{n x+1}$ is continuous on $[0,\infty)$ and monotonically increasing. So on the measurable space $\bigl((0,\infty),\mathcal{B}\bigl((0,\infty)\bigr)\bigr)$ there are measures with $$\mu_n\bigl((a, b]\bigl) = F_n(b) - F_n(a)\, .$$ These are finite measures, because $F_n(x) < 1$. It seems that the following limit exists for all $b > a$: $$\mu\bigl((a,b]\bigr ) := \lim_{n\rightarrow\infty} \mu_n\bigl((a,b]\bigr)\,.$$

Let $A_m:=\bigl(0,\frac{1}{m}\bigr)$, so $A_m\downarrow\emptyset$. Then $$\lim_{m\rightarrow\infty} \mu(A_m) = \lim_{m\rightarrow \infty} \lim_{n\rightarrow\infty} \mu_n(A_m) =\lim_{m\rightarrow \infty} \lim_{n\rightarrow\infty} \frac{n}{n+m} = 1\, .$$ But if $\mu$ were a measure, it would be monotonous.

Can you help me to see what exactly goes wrong here? Why can't the theorem be applied? Thank you!


Notice that in the theorem, Exercise 1.3.3 from Klenke's 2nd edition, you need to show that $$\mu_n(A)\to\mu(A)\ \ \ \ \mathbf{\text{ for all }}\ \ \ A\in\mathcal{A}$$ But we have only seen $\mu_n(A)$ converging for $A=(a,b]$.

$$\mu_n((a,b])\to\begin{cases}0&\text{ if }a>0\\1&\text{ if }a=0\end{cases}$$

When checking the convergence for other sets we will run into problems. The problem is that the convergence above is not uniform.

Observe that if $A_k=\left(\frac{1}{k+1},\frac{1}{k}\right]$ then $$\sum_{k=1}^{\infty}\mu(A_k)=\sum_{k=1}^{\infty}0=0<1=\mu((0,1])=\mu\left(\bigcup_{k=1}^{\infty}A_k\right)$$

Here there is a clever construction of a set $B$ formed by the union of some of the $A_n$ such that $\mu_n(B)$ is not convergent.

  • $\begingroup$ Setwise convergence for every set of a sigma-algebra is much more than convergence for a generating set of the sigma-algebra. It is a very powerful (and rare) phenomenon. $\endgroup$ – GEdgar Jan 18 '15 at 18:20
  • $\begingroup$ @GEdgar Well, some uniformity in the convergence would be enough to prove it follows. $\endgroup$ – Pp.. Jan 18 '15 at 18:25

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