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I cannot write a neat proof of this result, so I would like to see how to be precise in these kinds of arguments.. Here is the problem

Let $I=[0,1]$ and let $f\colon I\times\mathbb R\to \mathbb R$ be a function such that

i) $f(\cdot,x)$ is measurable for all $x\in\mathbb R$;

ii) $f(t,\cdot)$ is continuous for a.e. $t\in I$.

Prove that, for every continuous function $x\colon I\to\mathbb R$, the function

$$g_x(t):=f(t,x(t))$$ is measurable.

Thank you for your kindness..

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2 Answers

up vote 3 down vote accepted

Fix $x\colon[0,1]\to \Bbb R$ a continuous function and define $$g_n(t):=f\left(t,\frac{\lfloor nx(t)\rfloor}n\right),$$
where $\lfloor\cdot\rfloor$ denotes the floor function, that is, the map which gives to a real number the largest integer which is smaller or equal than this real number.

  • We check that $g_n$ is measurable, using the assumption i), and writing $g_n$ as as $$\lim_{k\to +\infty}\sum_{j=-k}^kf(t,j/n)\chi_{[j,j+1)}(nx(t)).$$ Indeed, the map $t\mapsto f(t,j/n)$ is measurable by i), and so is $t\mapsto f(t,j/n)\chi_{[j,j+1)}(nx(t))$. A sum of measurable functions is measurable, and pointwise limit of measurable functions still is measurable.
  • Then using ii), show that $g_n(t)$ to $g(t)$ for almost every $t$.
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Can you please provide more details in the first point of your argument? Still I cannot follow you completely even though you are convincing me you are right... Moreover, I can't see what is $j$ in the formula.. Of course later i will accept your answer.. Thank you Davide –  guido giuliani Jul 21 '12 at 14:43
    
It was a typo that I will correct readily. I didn't gave all the details since it's a homework question. –  Davide Giraudo Jul 21 '12 at 14:48
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Here is a less slick approach:

Suppose $x= \alpha 1_A$, with $A$ measurable. Then $g_x^{-1} V = \{t \in A|f(t,\alpha) \in V\} \cup \{t \in A^C|f(t,0) \in V\}$, hence $g_x$ is measurable.

Now let x be simple, $x = \sum_k \alpha_k 1_{A_k}$, where we may take the $A_k$ to be disjoint. Then we can write $g_x(t) = f(t,0)+\sum_k (f(t,\alpha_k 1_{A_k})-f(t,0))$, the sum of measurable functions, hence $g_x$ is measurable.

Now let $x$ be measurable, and let $s_n$ be a sequence of measurable simple functions such that $s_n(t)\to x(t)$ and $|s_n(t)| \leq |x(t)|$. Then $g_x(t) = \lim_{n \to \infty} g_{s_n}(t)$, so $g_x$ is the limit of measurable functions. Hence $g_x$ is measurable.

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