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I am studying Folland's Real analysis and I am stuck at following exercise (not a homework).

Suppose that $f$ is a function on $\mathbb{R} \times \mathbb{R}^k$ such that $f(x,\cdot)$ is Borel measurable for each $x \in \mathbb{R}$ and $f(\cdot,y)$ is continuous for each $y \in \mathbb{R}^k$. For $n \in \mathbb{N}$, define $f_n$ as follows. For $i \in \mathbb{Z}$ let $a_i = i/n$, and for $a_i \leq x \leq a_{i+1}$ let, $$f_n(x,y) = \frac{f(a_{i+1},y)(x-a_i)-f(a_i,y)(x-a_{i+1})}{a_{i+1}-a_i}$$ Then $f_n$ is Borel measurable on $\mathbb{R} \times \mathbb{R}^k$ and $f_n \to f$ pointwise; hence $f$ is Borel measurable on $\mathbb{R} \times \mathbb{R}^k$. Conclude by induction that every function on $\mathbb{R}^n$ that is continuous in each variable separately is Borel measurable.

Statement says $f_n$ is Borel measurable, by saying that it states it is Borel measurable for not only $f_n(x,\cdot)$ but also other variables of the function. I found this confusing.

Thanks in advance!

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Do you agree that for a constant $c \in \mathbb{R}$ and $f$ as in your question the function $g(x,y) = f(c,y) \cdot x$ is Borel measurable in both $x$ and $y$? – Martin Jan 11 '13 at 0:12
Since $y \in \mathbb{R}^k$ is continuous, yes. – oeda Jan 11 '13 at 0:17
I'm not sure what you mean by "since $y \in \mathbb{R}^k$ is continuous" ($f$ is only Borel measurable in the second argument). Anyway: the function $f_n$ is on each interval $(a_i,a_{i+1})$ a combination of two functions of the form $f(c,y) \cdot (x-d)$ (with $c$ and $d$ constant) and divided by another constant. – Martin Jan 11 '13 at 0:20
OK, I guess I got what you said. My previous comment is wrong clearly, continuity implies Borel measurability however converse is not true. For your last comment: Every function type of $f(c,y)$ is Borel measurable and we know (for example) if functions $m,n$ are Borel measurable, then $m \cdot n$ and $m+n$. I think your last comment is about that. Then we can deduce that $f_n$ is Borel measurable since it is the combination of the measurable functions. Am I right so far? – oeda Jan 11 '13 at 0:30
Yes, I think it would be a good idea to move to pointwise convergence. Keep in mind that the formula you wrote for $f_n$ holds on the pieces $[a_i,a_{i+1}] \times \mathbb{R}^k$. We argued that on each of these pieces we have a Borel measurable function, so $f_n$ is indeed measurable on $\mathbb{R} \times \mathbb{R}^k$. – Martin Jan 11 '13 at 1:01
up vote 3 down vote accepted

We already discussed why $f_n$ is measurable on $\left[\frac in,\frac{i+1}n\right]\times \mathbb{R}^k$. For $(x,y) \in \left[\frac in,\frac{i+1}n\right] \times \mathbb{R}^k$ we have $$ f_n(x,y) = \frac{f\left(\frac{i+1}{n},y\right)\cdot\left(\vphantom{y}x-\frac in\right)-f\left(\frac in,y\right)\cdot\left(x-\frac{i+1}n\right)}{\frac1n}. $$ In the numerator there is the difference of two products of measurable functions, hence it is measurable on $\left[\frac in,\frac{i+1}n\right]\times \mathbb{R}^n$. Using the partition $$\mathbb{R} \times \mathbb{R}^k = \bigcup_{i \in \mathbb{Z}} \left[\tfrac in,\tfrac{i+1}n\right]\times \mathbb{R}^k$$ one then sees that $f_n$ is measurable on all of $\mathbb{R} \times \mathbb{R}^k$.

To understand the definition of $f_n$, notice that for fixed $y \in \mathbb{R}^k$ the function $x \mapsto f_n(x,y)$ is continuous and it is linearly interpolating the values $f\left(\frac in,y\right)$ for $i \in \mathbb{Z}$: if $x = \frac{i}{n}$ then $f_n\left(\frac in,y\right) = f\left(\frac in,y\right)$. Once you observe this, pointwise convergence is not too hard to prove, using continuity of $x \mapsto f(x,y)$.

Concerning the induction in the last sentence of the exercise, suppose $n =2$ and that $f\colon \mathbb{R}^2 \to \mathbb{R}$ is continuous in each variable separately. Then $f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ satisfies the hypothesis of the exercise (with $k=1$) and it follows that it is measurable. For the induction step, assume that for $n$ it is already proved that a function on $\mathbb{R}^n$ which is continuous in each variable is measurable. Now consider a function $f \colon \mathbb{R}^{n+1} \to \mathbb{R}$ which is continuous in each variable. Interpreting $f$ as a function $f \colon \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}$, the induction hypothesis applies to $y \mapsto f(x,y)$ to show that it is Borel measurable and the exercise completes the induction step.

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This is not so much an answer, but rather an attempt at explaining what is going on. It seems that this is more of an obstacle than the actual technical details. If I misjudged, my apologies. – Martin Jan 13 '13 at 14:03
many thanks for your helps! – oeda Jan 13 '13 at 14:48

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