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Let $f$ be defined on $\mathbb{R}^2$ such that for all $a \in \mathbb{R}$ the function $y \mapsto f(a,y)$ is Borel-measurable and such that for all $b \in \mathbb{R}$ the function $x \mapsto f(x,b)$ is continuous.

I want to show that for all $a,b,c \in \mathbb{R}$ the function $(x,y) \mapsto bx + c f(a,y)$ is Borel-measurable on $\mathbb{R}^2$.

The only thing I can come up with, is that the combination of Borel-measurable/continuous functions is again Borel-measurable/continuous. However, I don't know if this is sufficient. Any ideas?

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    $\begingroup$ It is sufficient as you say. $\endgroup$ – John B Dec 27 '16 at 21:17
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Your function is the sum of the continuous (hence Borel measurable) function $(x,y)\mapsto bx$ and the Borel measurable function $(x,y)\mapsto y\mapsto f(a,y)$; this sum is therefore Borel measurable.

More is true: $f$ is itself Borel measurable, being the pointwise limit of the sequence $$ f_n(x,y):=\sum_{k\in\Bbb Z} f(k2^{-n},y)1_{[k2^{-n},(k+1)2^{-n})}(x). $$

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