# A Dominated Convergence Theorem for Product Measurable Functions

Let $$(X,\mathcal X,\mu)$$ and $$(Y,\mathcal Y,\nu)$$ be finite measure spaces. We denote the product $$\sigma$$-algebra by $$\mathcal X \otimes \mathcal Y$$. Let $$\{ f_n \}$$ be a sequence of functions where each $$f_n\colon X\times Y \to \mathbb R$$ is $$\mathcal X \otimes \mathcal Y$$-measurable.

Suppose that for each $$y\in Y$$, $$f_n(\cdot,y)$$ converges in $$\mu$$-measure to $$f(\cdot,y)$$. We may also assume that $$\vert f_n(x,y)\vert \le G(x,y)$$, where $$G$$ is $$\nu$$-integrable for $$\mu$$-almost all $$x$$. Do we then have that, for some appropriate notion of convergence, $$\lim_{n\to \infty} \int_Y f_n (\cdot,y) \, \mathrm d\nu = \int_Y f(\cdot,y)\,\mathrm d \nu? \tag{1}\label{1}$$

By convergence in $$\mu$$-measure for each $$y$$, I mean that, for each $$\varepsilon > 0$$ and each $$y \in Y$$, $$\lim_{n\to \infty} \mu \left( \left\{ x: \vert f_n (x,y) - f(x,y) \vert > \varepsilon \right\} \right) = 0.$$

Some Thoughts

If for each $$y$$, $$f_n(x,y)\to f(x,y)$$ for $$\mu$$-almost all $$x$$, then \eqref{1} holds, where the limit is taken for $$\mu$$-almost all $$x$$. This is an easy consequence of the Dominated Convergence Theorem.

If $$f_n(\cdot,y)\to f(\cdot,y)$$ in $$\mu$$-measure for each $$y$$, then one might consider passing to a subsequence that converges for $$\mu$$-almost all $$x$$. However, this subsequence now depends on $$y$$, so it is not clear how to apply the previous argument to this case.

If instead $$f_n(\cdot,y) \to f(\cdot,y)$$ in $$\mu$$-measure uniformly in $$y$$ (that is, $$\sup_y \vert f_n(\cdot,y) - f(\cdot,y) \vert \to 0$$ in $$\mu$$-measure), then \eqref{1} also holds, taking limits in $$\mu$$-measure. To see this, note that $$\mu \left( \left\{ x: \left\vert \int_Y (f_n (x,y) - f(x,y))\,\mathrm d \nu \right\vert > \varepsilon \right\} \right) \le \mu \left( \left\{ x: \sup_y \left\vert f_n (x,y) - f(x,y)\right\vert\ > \frac{\varepsilon}{\nu(Y)} \right\} \right)$$ where the right-hand side goes to $$0$$ as $$n \to \infty$$. (The finiteness of $$\nu$$ is important here.)

Does anyone have any suggestions for the general case?

I claim that your convergence in (1) holds in measure. To prove this, define $$F_n (x) := \int_Y f_n(x,y) \, d \nu(y)$$, and define $$F$$ similarly. To prove the convergence in measure, it suffices to show that for each subsequence $$(F_{n_k})_k$$, there is a further subsequence $$(F_{n_{k_\ell}})_\ell$$ which converges almost everywhere to $$F$$. For convenience, I will not bother with the first subsequence, and instead work with the sequence $$(F_n)_n$$ itself.
First, I claim that we actually have $$f_n \to f$$ in measure on $$X \times Y$$. To see this, note by Fubini's theorem that $$(\mu \otimes \nu) (\{ (x,y) : |f_n (x,y) - f(x,y)| > \epsilon \}) = \int_Y \mu (\{ x \in X : |f_n(x,y) - f(x,y)| > \epsilon \}) \, d \nu(y).$$ By your assumption, the integrand converges pointwise to zero. Furthermore, since $$\mu,\nu$$ are finite measures, the integrand is dominated by $$y \mapsto \mu(X)$$, which is integrable over $$Y$$. Hence, we get $$(\mu \otimes \nu) (\{ (x,y) : |f_n (x,y) - f(x,y)| > \epsilon \}) \to 0$$ by dominated convergence.
Now, since $$f_n \to f$$ in measure, there is a subsequence $$(f_{n_k})_k$$ such that $$f_{n_k} \to f$$ almost everywhere. By an application of Fubini's theorem, this implies that there is a null-set $$N \subset X$$ such that for each $$x \in X \setminus N$$, there is a null-set $$N_x \subset Y$$ satisfying $$f_{n_k} (x,y) \to f(x,y)$$ for all $$y \in Y \setminus N_x$$. Since we also have $$|f_{n_k} (x,y)| \leq G(x,y)$$ (where by enlarging $$N$$, we can assume that $$G(x,\cdot)$$ is integrable), the dominated convergence theorem shows $$F_{n_k}(x) = \int_Y f_{n_k}(x,y) \, d \nu(y) \to \int_Y f(x,y) \, d \nu(y) = F(x)$$. This holds for all $$x \in X \setminus N$$, and hence almost everywhere.
• Very nice! Thank you. I just have one point to check to make sure I understand. Am I correct in thinking the following? Let $M \subset X\times Y$ be the null set on which $(f_{n_k})_k$ does not converge. Then, we can write $N_x = \{ y\in Y : (x,y) \in M\}$ and $N = \{ x \in X : (x,y)\in M\;\text{for some}\;y\}$. Or am I mistaken? Oct 8, 2019 at 5:52
• @Theoretical: The choice of $N$ is not quite right, I think. The argument I had in mind is that $0=(\mu\otimes\nu)(M)=\int_X \nu(N_x)\,d\mu(x)$, where $N_x$ is as suggested by you. Now, if the integral over a non-negative function vanishes, the integrand vanishes almost everywhere, so you can choose $N = \{x : \nu(N_x) \neq 0\}$, and this is indeed a null-set. Oct 8, 2019 at 6:30