Let $X$ be a measure space. Let $S\_j$, $j \in \mathbb N$ be an increasing sequence of $\sigma$-algebras on $X$ such that $S := \bigcup\_{j \geq 0} S\_j$ is a $\sigma$-algebra. For every $j$, let $\mu\_j$ be a probability measure on $S\_j$.


Let $f\_{ij}$ $(i, j \in \mathbb N)$ be a double indexed sequence of functions such that that for every $j$, $f\_{ij}$ converges $\mu\_j$-a.e. to a $S\_j$ measurable function $f\_j$.


Suppose there exists some probability measure $\mu$ on $S$ such that $f\_j$ converges $\mu$-a.e. to a function $f$.


Suppose further that:


* the restriction of $\mu$ to $S\_j$ is absolutely continuous with respect to $\mu\_j$ for every $j$
* $f\_{ij}, f\_j, f$ are $\mu$-integrable, and $f\_{ij}$ is $\mu\_j$-integrable for every $i, j$
* $\int f\_{ij} \, d \mu\_j \to \int f\_j \, d \mu\_j$ for every $j$
* $\int f\_j \, d \mu \to \int f \, d \mu$


Is it true that there exists a increasing function $b: \mathbb N \to \mathbb N$ such that $\int f\_{n, b(n)} \, d \mu$ converges to $\int f \, d \mu$?



Measure theory problem concerning convergence of integrals
Following Iosif's comment: the family $\sum\_j S\_j$ virtually never is a $\sigma$-algebra.


But this does not matter: even if $S = S\_j$ for all $j$, the claim is clearly false without further restrictions. Indeed, consider $[-1,1]$ with the usual Borel $\sigma$-algebra $S = S\_j$, and
$$
\begin{gathered}
 \mu(dx) = \tfrac{1}{2} (1 + x) dx , & \mu\_j(dx) = \tfrac{1}{2} dx , \\
 f(x) = f\_j(x) = 0 , & f\_{ij}(x) = 2 i x^{2 i - 1} .
\end{gathered}
$$
Then:
$$
\begin{gathered}
 \lim\_{i \to \infty} f\_{ij}(x) = f\_j(x) \text{ except when $x = \pm 1$,} \\
 f\_j(x) = f(x) \text{ for all $x$,} \\
 \int f\_{ij}(x) \mu\_j(dx) = 0 = \int f\_j(x) \mu\_j(dx) , \\
 \int f\_j(x) \mu(dx) = 0 = \int f(x) \mu(dx) ,
\end{gathered}
$$
but neither of:
$$
\begin{gathered}
 \int f\_{i,j(i)}(x) \mu(dx) = \frac{2 i}{2 i + 1} \, , \\
 \int f\_{i(j),j}(x) \mu(dx) = \frac{2 i(j)}{2 i(j) + 1}
\end{gathered}
$$
can converge to $0 = \int f(x) \mu(dx)$.


