Csdrhrt

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

Suppose also that there exists a measure $mu$ such that $mu_k(A)tomu(A)$ for every $Ainmathcal F$.
Can we conclude that $mu=flambda$ for some R.-N. derivative $f$, and that $f_kto f$ 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!

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
$$
f = frac{dmu}{dlambda}
$$ 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
$$
int_A g(x) dlambda(x) = lim_{rtoinfty} int_A f_{n(k(r))}(x) dlambda(x) =lim_{rtoinfty}mu_{n(k(r))}(A) = mu(A) = int_A f(x)dlambda(x),
$$ for all $Ainmathcal{F}$. This shows $g =f$. Since every subsequence admits a convergent subsubsequence with common limit $f$, we know that $$lim_{ntoinfty} f_n = f$$ 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 $f$.

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 $[frac {i-1} {2^{n}},frac i {2^{n}})$, $1leq i leq 2^{n}, ngeq 1$ 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 $mu_n (A)=int_A g_n dlambda$. Then each $mu_n$ is a probability measure, $mu_n << lambda$ adn $dmu_n= g_n dlambda$. Note that $0leq g_n leq 2$. 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].