Let $X$ be a metric space, and let $\mu_n$ be a sequence of measures on $X$ converging weakly to a measure $\mu$, meaning for all bounded continuous functions $f$, we have $\int_{X}fd\mu_n \rightarrow \int_{X}fd\mu$ as $n \to \infty$.

Now say that $f$ is not necessarily bounded, but it is continuous, and that $\sup_n \int_{X}fd\mu_n < \infty$ and $\int_X fd\mu < \infty$. Is it necessary that $\int_X fd\mu_n \rightarrow \int_Xfd\mu$?

What I tried: Assume further that $f$ is non-negative and that we can get a sequence of functions $f_n$ with $f_n \leq f_{n+1}$ and converging pointwise to $f$, which are bounded continuous. Then using monotone convergence + weak convergence we get

$$\int_X f d\mu = \lim_{m\to\infty}\int_X f_m d\mu = \lim_{m\to\infty}\lim_{n\to\infty}\int_X f_m d\mu_n$$

But I'm not sure how to proceed from here. When can one interchange these limits?


No, this fails especially when $\mu$ is atomic. Take $\mu_n(\{0\})=1-1/n$ and $\mu_n(\{n\})=1/n$. Then $\mu_n$ converges weakly to the point mass at 0. Now take $f(x)=x$. Then $\sup_n \int_X fd\mu_n=1$ but $\int_X fd\mu_n=1\neq 0=\int_Xfd\mu$.

Usually we speak of measures losing mass in the limit, however in this case it's a loss of expectation for unbounded functions.

  • $\begingroup$ This example is interesting. The mass is concentrating around 0 but the support is also spreading out. I wonder if the result holds when convergence of the first moment is added. $\endgroup$ – ttb Jun 24 '16 at 22:59
  • 1
    $\begingroup$ @user65308: Nope, modify the above to $\mu(0)=1-1/n$ and $\mu(\pm n)=1/2n$. Then the first moment converges. However try $f(x)=x^2$. $\endgroup$ – Alex R. Jun 24 '16 at 23:48

A small remark: One can prove that $$ \liminf_{n \to \infty} \int_X f d\mu_n \geq \int_X f d\mu. $$ The proof follows essentially from Approximating integrals of continuous functions..


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