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Suppose $f$(x,y) is a function on $\mathbb{R}^{2}$ that is separately continuous: for each fixed variable, $f$ is continuous in the other variable. Prove that $f$ is measurable on $\mathbb{R}^{2}$.

There is also a hint: Approximate $f$ in the variable $x$ by piecewise-linear functions $f_n$ so that $f_n$ $\rightarrow$ $f$ pointwise.

I don't get how to prove the problem via this hint, or is there any other approach?

Rudy the Reindeer
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Alex
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    It is actually sufficient that the function is continuous in one coordinate and measurable in the other. Such functions are known as Caratheodory functions. – Michael Greinecker Dec 28 '12 at 17:36

2 Answers2

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Hint: take $f_{m,n}(x,y):=f\left(\frac{\lfloor mx\rfloor}n,\frac{\lfloor ny\rfloor}n\right)$, where $\lfloor t\rfloor$ is the greatest integer lower than $t$. As $f_{m,n}$ takes only countably many values, it's measurable.

Davide Giraudo
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  • why does $f_{m,n}$ converges to $f$, and how is continuity used here? – Alex Dec 28 '12 at 17:45
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    For a fixed $m$, $f_{m,n}(x,y)$ converges to $f\left(\frac{\lfloor mx\rfloor}m,y\right)$ as the map $t\mapsto f\left(\frac{\lfloor mx\rfloor}m,t\right)$ is continuous. Do the same taking the limit with respect to $m$. – Davide Giraudo Dec 28 '12 at 17:47
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    @DavideGiraudo I still have difficulty understanding the convergence. Can you help on my post http://math.stackexchange.com/questions/1388414/show-lim-m-to-infty-n-to-infty-f-frac-left-lfloor-mx-right-rfloo? Thank you! – Tony Aug 08 '15 at 03:01
  • Wait, so this answer does not work, if I understand the other question correctly? – Bart Michels Jul 14 '18 at 11:53
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Inspired by Cardona's answer: Showing a function of two variables is measurable, I share my answer:

Construct $f_n(x,y)=f(k/n,y)$, where $x\in[k/n,(k+1)/n$.It can be proved that $lim_{n\to\infty}f_n(x,y)=f(x,y)$. Applying the property of measurable function: f(x,y) is mearsurable if $f_n(x,y)$ is measurable for each n, that is $\forall a, \{f_n>a\}$ is measurable.

Note that $\{f_n > a\} = \{(x, y) : f_n(x, y) > a\} = \bigcup_k \Bigg\{ \bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg) \times \{y : f\bigg( \frac{k}{n}, y \bigg) > a\}\Bigg\}$.

Obviously $\bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg)$ is measurable. And for $\{y : f\bigg( \frac{k}{n}, y \bigg) > a\}$, since $f(x,y)$ is continuous in $y$ for a fixed $x=k/n$ and $(a,+\infty)$ is open, $\{y : f\bigg( \frac{k}{n}, y \bigg) > a\}$ is open, therefore is measurable. Hence $\{f_n > a\} = \bigcup_k \Bigg\{ \bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg) \times \{y : f\bigg( \frac{k}{n}, y \bigg) > a\}\Bigg\}$ is measurable.

The proof of $lim_{n\to\infty}f_n(x,y)=f(x,y)$:

$|f_n(x,y)-f(x,y)|=|f(k/n,y)-f(x,y)|$, where $x\in\bigg[\frac{k}{n}, \frac{k + 1}{n} \bigg)$. Since f is continuous on x for a fixed y(applying the separate continuity second time), $\forall\epsilon>0, \exists n s.t.\delta=1/n, $ when $x-k/n<\delta=1/n, |f(x,y)-f(k/n,y)|<\epsilon$, that is $lim_{n\to\infty}f_n(x,y)=f(x,y)$.

See Cardona's answer for a similar question: Showing a function of two variables is measurable for more details

EnHu
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