# Measure theory problem concerning convergence of integrals

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 $$\int f \, d \mu$$?

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