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!


1 Answer 1


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
    Commented Jan 18, 2015 at 18:20
  • $\begingroup$ @GEdgar Well, some uniformity in the convergence would be enough to prove it follows. $\endgroup$
    – Pp..
    Commented Jan 18, 2015 at 18:25
  • $\begingroup$ @Pp.. does the limit existence here is valid with a value equal to the $+ \infty$ element ? $\endgroup$ Commented Jul 6, 2023 at 20:01

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