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Please help me to prove the following;

Let $ \ f:[0,1]^2\longrightarrow\mathbb{R}$ be such that:

(i) $\ f(x,\cdot)$ is measurable for each fixed $x\in[0,1]$;

(ii) $\ f(\cdot,y)$ is continuous for each fixed $y\in[0,1]$.

Show that $\ f$ is measurable.

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    $\begingroup$ Can you share what you've tried, and explain what you're having trouble with? $\endgroup$ – user61527 Feb 2 '14 at 18:58
  • $\begingroup$ math.stackexchange.com/questions/659305/… This is what I have. But it seems incorrect. $\endgroup$ – Guest_000 Feb 2 '14 at 19:00
  • $\begingroup$ See here. $\endgroup$ – Michael Greinecker Feb 2 '14 at 19:06
  • $\begingroup$ The question that this question is a duplicate of has since been deleted; please reopen. $\endgroup$ – Eric Stucky Jan 3 '15 at 22:43
  • $\begingroup$ Regarding the answers, it seems that we only need left/right continuity of $f(\cdot, y)$. $\endgroup$ – Junnan Feb 13 '18 at 9:55
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Suppose $f(x, y)$ is continuous in $x$ and measurable in $y$. Let $f_n: \mathbb{R}^2 \rightarrow \mathbb{R}$ be defined by $f_n(x, y) = f(m/n, y)$ where $m, n$ are integers $n \geq 1$ and such that $(m-1)/n \leq x < m/n$. Then $f_n$'s are all measurable. Now observe that, since $f$ is continuous in $x$, for fixed $(x, y)$, as $n \rightarrow \infty$, $f_n(x, y) \rightarrow f(x, y)$.

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  • $\begingroup$ One question, is the measurabilty here (because the source of the question doesn't specify) is under Borel measure or Lebesgue measure? $\endgroup$ – Guest_000 Feb 3 '14 at 18:23
  • $\begingroup$ Doesn't matter which one. Both classes are closed under pointwise limits. $\endgroup$ – hot_queen Feb 4 '14 at 1:06
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This sort of problem frequently boils down to approximating over simple functions of intervals.

In this problem, one approach is as follows: Let $\eta_n(t) = {\lfloor n t \rfloor \over n}$. Then for $k \in \mathbb{Z}$, we have $\eta_n(t) = {k \over n}$ iff $t \in [ k, k+ { 1 \over n})$. Then we can 'approximate' $f$ by $f(\eta_n(x),y)$ and use the facts that $\eta_n$ is constant over $[ k, k+ { 1 \over n})$ and $\lim_n \eta_n(t) = t$.

Let $\phi_n(x,t) = f(\eta_n(x),y) = \sum_{k \in \mathbb{Z}} 1_{[ k, k+ { 1 \over n})} (x) f({k \over n}, y)$. Since $(x,y) \mapsto 1_{[ k, k+ { 1 \over n})} (x)$ and $(x,y) \mapsto f({k \over n}, y)$ are measurable, we see that the $\phi_n$ are measurable.

Continuity gives $\phi_n(x,y) \to f(x,y)$, hence $f$ is measurable.

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