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 $\int f \, d \mu$?

  • $\begingroup$ $S := \cup_{j \geq 0} S_j$ is not necessarily a sigma algebra. So, what do you then mean by $\mu$ and $\int f\,d\mu$? $\endgroup$ – Iosif Pinelis May 22 '19 at 16:25
  • $\begingroup$ Ah let me modify this.. $\endgroup$ – James Baxter May 22 '19 at 16:32

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 $0 = \int f(x) \mu(dx)$.

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    $\begingroup$ You beat me to it. The strategy is: take any example of strict inequality in Fatou's lemma, and you are practically done. $\endgroup$ – Gerald Edgar May 23 '19 at 0:04

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