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Given $f(\cdot, y)$ is measurable for each $y$, $f(x, \cdot)$ is continuous for each $x$. If $u(t)$ is continuous, how can I show that the function $f:[0,1]\times \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(t, u(t))$ is measurable?

I know the standard way to do this is to show that for each $B\in \mathcal{B}(\mathbb{R})$ we have $$f^{-1}(B) \in \mathcal{B} ([0,1]\times \mathbb{R}).$$

But I am having trouble writing down the pre-image explicitly, could anyone give me a hint. Thank you very much!

Xiao
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  • You are using $f$ about two different functions. In particular, $f(t,u(t))$ is a function of a single variable, not two variables as you indicated, Rename one of the $f$s and clean up your question, and see if that doesn't help towards finding the answer. – Harald Hanche-Olsen Sep 10 '14 at 07:56
  • open sets in $\mathbb{R}^2$ can be written as arbitrary union of $A\times B$ where $A$ and $B$ are both open in the $\mathbb{R}$ topology? – Xiao Sep 10 '14 at 07:56
  • You can replace “arbitrary” by “countable” if you wish. – Harald Hanche-Olsen Sep 10 '14 at 07:57

1 Answers1

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More generally you have :

If $u,v$ are real measurable functions on a measurable space $X$ and $\Phi$ be a continuous mapping from plane into a topological space $Y$. If we define $h(x)=\Phi(u(x),v(x))$ then $h$ is measurable.

Set $h=\Phi\circ f$ where $f:\mathbb{R}\rightarrow \mathbb{R}\times \mathbb{R}$ with $x\rightarrow (u(x),v(x))$

See that basic open set in $\mathbb{R}\times \mathbb{R}$ looks like $I_1\times I_2$ where $I_1,I_2$ are open in $\mathbb{R}$.

Now, Prove that $f^{-1}(I_1\times I_2)=u^{-1}(I_1)\cap v^{-1}(I_2)$

Assuming that $u,v$ are measurable should tell you that $f$ is measurable...

Now see that composition of continuous and measurable is measurable...

So, $h$ is measurable and this is what we want..