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.

  • $\begingroup$ 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? $\endgroup$ – Theoretical Economist Oct 8 '19 at 5:52
  • $\begingroup$ @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. $\endgroup$ – PhoemueX Oct 8 '19 at 6:30

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