I want to relate strong convergence of measures and usual notions of convergence for functions. Let $(X,\mathcal{A})$ be a measurable space

Definition: A sequence of probability measures $\mu_n$ converges strongly to $\mu$ if $\mu_n(A)\to\mu(A)$ for all $A\in\mathcal{A}$.

Now suppose $\mu$ is a probability measure, $f_n,f$ are nonnegative measurable functions with $\int_X f_nd\mu=\int_X fd\mu=1$. Set $\nu_n(A)=\int_Af_nd\mu$ and $\nu(A)=\int_A fd\mu$.

Strong convergence of $\nu_n\to \nu$ becomes as follows: $\int_A f_nd\mu\to\int_A fd\mu$ for all $A\in\mathcal{A}$.

Question: If $\nu_n\to \nu$ strongly, does $f_n\to f$ pointwise a.e.? In measure? Almost uniformly? In any other sense?

We can also state everything "backwards", in an equivalent manner: If $\nu_n,\nu$ are given probability measures, set $\mu=\nu/2+\sum_{n=1}^\infty 2^{-n-1}\nu_n$ and $f_n=d\nu_n/d\mu$, $f=d\nu/d\mu$. The same question above applies.

All I could get with Fatou's Lemma is that $f\geq\liminf f_n$, which doesn't seem really useful.

  • $\begingroup$ This is far from an answer, but convergence in $L^1$ certainly suffices since $$ |\nu_n(A) - \nu(A)| = \left|\int_Af_n-fd\mu\right| \leq \int_A|f_n-f|d\mu \leq \int_X|f_n-f|d\mu \xrightarrow{n \to\infty} 0 $$ $\endgroup$ – Guido A. Jun 27 '18 at 5:48

In general you won't have any type of pointwise convergence or convergence in measure.

As a counterexample consider $f \equiv 1$ on $[0,1]$ and $f_n (x) = 1+\cos(2\pi n x)$. It is then not hard to see $\int f_n g d\mu \to \int f g d \mu$ for all $g \in L^2$, where I am considering the Lebesgue measure. To see this, use the density of $C_c^\infty$ in $L^2$, and for a $C_c^\infty$ function use partial integration.

Now, note $\int |f-f_n| d\mu \geq c >0$. But if we would have $f_n \to f$ in measure, then the dominated convergence would imply $\int |f-f_n|d\mu \to 0$.

Finally, since the indicator functions span a dense subset of $L^\infty$, your convergence is equivalent to $f_n \to f$ in the weak sense in $L^1$ (this also uses that $\|f_n\|_{L^1}$ is bounded).


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