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Let f(x,y) be a function defined on the unit square $0\leq x\leq1$, $0\leq y\leq1$ that is continuous on each variable separately. Is f a measurable function of (x,y)?

I think I need to look at the pre-images of f, and I need to use the fact that it is continuous. Maybe I can use the epsilon-delta definition of continuous functions?

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Satisfied with an answer below? – Did Oct 25 '12 at 11:46

The answer is YES. For a proof, see Separate Continuity and Measurability, B. E. Johnson, Proceedings of the American Mathematical Society Vol. 20, No. 2 (1969), pp. 420-422.

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Theorem: Let $f(x,y)$ be a function defined on the unit square $0\le x\le 1,0\le y\le 1$ which is continuous in each variable separately. Show that $f$ is a measurable function of $(x,y).$

Proof: If we have a sequence of measurable functions $f_n$ such that $f_n \to f$, then $f$ is measurable.

Define $f_n(x, y) = f_n\bigg(x, \frac{k}{n} \bigg)$ where $\frac{k}{n} \leq y < \frac{k + 1}{n}$. We are partitioning the $y$-axis unit interval into $n$ equal partitions and then looking at the largest endpoint of one of those partitions such that $y$ is greater than it, e.g., say $n$ = 3, then we have the following partition points $\{0, 1/3, 2/3, 1\}$, if we chose $y = 1/2$, then the largest endpoint such that $y$ is greater than it would be $1/3$. By taking $n$ to be arbitrarily large, we are squeezing $y$ into to rational numbers and looking at the lower one. Note that $f_n(x, y) \to f(x, y)$ since for $(x_0, y_0) \in X \times Y$, $\vert f(x_0, y_0) - f_n(x_0, y_0) \vert = \vert f(x_0, y_0) - f(x_0, k/n) \vert$, but as $n \to \infty$, $k/n \to y_0$ and since $x_0$ is fixed in this case, we can use the continuity of $f$ in the $y$ variable to show that $f(x_0, k/n) \to f(x_0, y_0)$.

We must show $f_n$ is measurable for all $n$, but that means for all finite $\alpha$ and $n \in \mathbb N$, $\{f_n > \alpha\}$ is measurable. Observe that $\{f_n > \alpha\} = \{(x, y) \in [0, 1] \times [0, 1] : f_n(x, y) > \alpha\} = \bigcup_{k = 0}^{n - 1} \Bigg\{ \{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\} \times \bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg) \Bigg\}$ but since $f$ is continuous in $y$, $\{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\} = \{f^{-1} \big( (\alpha, \infty) \big)$ is open and thus measurable and $\bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg)$ is an interval, so it is trivially measurable. Recall that the Cartesian product of two measurable sets is measurable. We are taking a finite union of measurable sets which is measurable, so we conclude $f_n$ is measurable for $n \in \mathbb N$.

The big question here, is: why is $\{f_n > \alpha\} = \bigcup_{k = 0}^{n - 1} \Bigg\{ \{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\} \times \bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg) \Bigg\}$ true? Observe the following

  • $(\subseteq)$ If $(x_0, y_0) \in \{f_n > \alpha\}$, then $f_n(x_0, y_0) = f_n\bigg(x_0, \frac{k}{n} \bigg) > \alpha$ where $\frac{k}{n} \leq y_0 < \frac{k + 1}{n}$ for some $k \in \{0, 1, \ldots, n - 1\}$. But then $x_0 \in \{x \in [0, 1] : f\bigg( x, \frac{k}{n} > \alpha\}$ and $y \in \bigg[ \frac{k}{n}, \frac{k + 1}{n} \bigg)$. so $(x_0, y_0) \in \bigcup_{k = 0}^{n - 1} \Bigg\{ \{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\} \times \bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg) \Bigg\}$.
  • $(\supseteq)$ If $(x_0, y_0) \in \bigcup_{k = 0}^{n - 1} \Bigg\{ \{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\} \times \bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg) \Bigg\}$, then there exists $k \in \{0, 1, \ldots, n - 1\}$ such that $(x_0, y_0) \in \{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\} \times \bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg)$. But then $x_0 \in \{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\}$ where $\frac{k}{n} \leq y_0 < \frac{k + 1}{n}$. But $x_0 \in \{x \in [0, 1] : f\bigg( x, \frac{k}{n}\bigg) > \alpha\} = \{ f_n > \alpha\}$.
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