Radon-Nikodym derivative of a limit of measures with bounded R.-N. derivative

Question

Let $(\Omega,\mathcal F)$ be a measurable space. Let $\lambda$ be a probability measure, and let $(\mu_k)_{k\geq 0}$ be a sequence of probability measures. Suppose that $\mu_k\ll\lambda$ for any $k$, say with R.-N. derivative $f_k$. Assume that there is a constant $C\geq 1$ such that each $f_k$ is bounded by $C$.

Suppose also that there exists a measure $\mu$ such that $\mu_k(A)\to\mu(A)$ for every $A\in\mathcal F$. Can we conclude that $\mu=f\lambda$ for some R.-N. derivative $f$, and that $f_k\to f$ almost everywhere (with respect to $\lambda$)?

I cannot find a proof for this statement, although it seems very believable to me...

A reference or a proof would be very much appreciated!

Answer 1

Have you heard about Vitali-Hahn-Saks theorem? If you have not, see https://en.wikipedia.org/wiki/Vitali%E2%80%93Hahn%E2%80%93Saks_theorem.

It says that a pointwise limit $\mu$ of sequence of measures $\mu_n$ which are absolutely continuous with respect to $\lambda$ is an absolutely continuous measure with respect to $\lambda$. Hence, the assumption that $\mu$ is a measure is not really necessary, and we know there exists the Radon-Nikodym derivative $$ f = \frac{d\mu}{d\lambda} $$ of $\mu$ with respect to $\lambda$. Observe that the family $\{f_n\}\subset L^1(\lambda)$ is uniformly bounded and thus is uniformly integrable. By Dunford-Pettis theorem, any subsequence $f_{n(k)}$ has a $L^1$-convergent subsubsequence $f_{n(k(r))}$ with limit $g$ (possibly depending on the subsequence). But then we must have $$ \int_A g(x) d\lambda(x) = \lim_{r\to\infty} \int_A f_{n(k(r))}(x) d\lambda(x) =\lim_{r\to\infty}\mu_{n(k(r))}(A) = \mu(A) = \int_A f(x)d\lambda(x), $$ for all $A\in\mathcal{F}$. This shows $g =f$. Since every subsequence admits a convergent subsubsequence with common limit $f$, we know that $$\lim_{n\to\infty} f_n = f$$ in $L^1(\lambda)$. However, pointwise convergence is not guaranteed as we can see from the general fact that there is a convergent $L^1$-sequence not converging pointwise (note that your assertion implies that any $L^1$-converging bounded sequence also converges almost everywhere). Of course, under additional assumptions, e.g. monotonicity of $\{f_n\}$, pointwise limit may exist and it must be equal to $f$.

Answer 2

Almost everywhere convergence need not hold. Here is a counterexample: Let $\lambda$ be Lebesgue measure on $(0,1)$. Arranging the indicator functions of the intervals $[\frac {i-1} {2^{n}},\frac i {2^{n}})$, $1\leq i \leq 2^{n}, n\geq 1$ in a sequence we get a sequence of measurable functions converging in measure but not converging at any point. Let $g_n=\frac {1+f_n} {1+\int_0^{1} f_n}$. Let $\mu_n (A)=\int_A g_n d\lambda$. Then each $\mu_n$ is a probability measure, $\mu_n << \lambda$ adn $d\mu_n= g_n d\lambda$. Note that $0\leq g_n \leq 2$. Now $\mu_n (A) \to \lambda (A)$ fro every Borel set $A$; this follows from the well known (and easy to prove) fact that almost everywhere convergence can be replaced by convergence n measure in DCT. However $g_n$ does not converge at any point! [ I am taking $\mu =\lambda$ here].