Is $g(x):=F(f_1(x),f_2(x),...,f_n(x))$ measurable where F is continuous and each f is measurable. Let $f_1,..., f_n: \mathbb{R}\rightarrow\mathbb{R}$ be measurable functions. And $F:\mathbb{R}^n\rightarrow\mathbb{R}$ be continuous. 
For $g:\mathbb{R}\rightarrow\mathbb{R}$, $g(x):=F(f_1(x),f_2(x),...,f_n(x))$. Is $g$ measurable?
I think I was be able to do that if F is measurable, but I don't know how is it if F is only continuous.
 A: It is well known that there exist continuous functions $f : [0,2] \to \mathbb{R}$ that are not Lebesgue-measurable. Extend a function of this to the rest of the real line continuously, via 
$$
h(x) = \cases{f(0) \quad x < 0 \\ f(x) \quad 0 \leq x \leq 2 \\ f(2) \quad x > 2}
$$
Now, take $n = 1$, $f_1 \equiv id$, $F \equiv h$.
A: I think I was confusing myself. All real continuous function should be Lebesgue measurable, which gives $F$ should be also measurable, which lead to $h$ is also measurable.
A: Assume:


*

*Let $\mathcal{X}$ be any set. 

*Let $\Sigma$ be any sigma-algebra on $\mathcal{X}$. 

*Fix $n$ as a positive integer. 

*For each $i \in \{1, ..., n\}$ and each $c \in \mathbb{R}$,  the function $f_i:\mathcal{X}\rightarrow\mathbb{R}$ has the property that $\{x \in \mathcal{X} : f_i(x)>c\}$ is in $\Sigma$. 

*The function $F:\mathbb{R}^n\rightarrow \mathbb{R}$ is continuous. 
Claim:
For each $c \in \mathbb{R}$, the set $\{x \in \mathcal{X} : F(f_1(x), ..., f_n(x)) > c\}$ is in $\Sigma$.  
Proof:
Fix $c \in \mathbb{R}$.  Define the set $\mathcal{Y} \subseteq\mathbb{R}^n$ by
$$ \mathcal{Y} = \{y \in \mathbb{R}^n : F(y) > c\}$$
Since $F$ is a continuous function, we know that $\mathcal{Y}$ is an open subset of $\mathbb{R}^n$. Hence, it can be written as the countable union of open boxes in $\mathbb{R}^n$: 
$$ \mathcal{Y} = \cup_{i=1}^{\infty} B_i $$
where each box $B_i$ has the form: 
$$ B_i = (a_{1i}, b_{1i}) \times (a_{2i}, b_{2i}) \times ... \times (a_{ni}, b_{ni}) $$
Then 
\begin{align}
\{x \in \mathcal{X} : F(f_1(x), ..., f_n(x)) > c\} &= \{x \in \mathcal{X} : (f_1(x), ..., f_n(x)) \in \mathcal{Y}\}\\
&= \cup_{i=1}^{\infty} \{x \in \mathcal{X} : (f_1(x), ..., f_n(x)) \in B_i\}\\
&= \cup_{i=1}^{\infty} \underbrace{\cap_{j=1}^n \underbrace{\{x \in \mathcal{X} : f_j(x) \in (a_{ji}, b_{ji})\}}_{\mbox{*in $\Sigma$}}}_{\mbox{**in $\Sigma$}} 
\end{align}
This is the countable union of sets in $\Sigma$  and hence is in $\Sigma$. $\Box$

*Note that for each $j \in \{1, ..., n\}$ and each open interval $(a,b)$ we have 
$$\{x \in \mathcal{X} : f_j(x) \in (a,b)\} \in \Sigma $$
because 
$$ \{x \in \mathcal{X} : f_j(x) \in (a,b)\} = \cup_{k=1}^{\infty} \left(\{x : f_j(x)>a\} \cap \{x : f_j(x)\leq b-1/k\}\right)$$
and this is a countable union of sets in $\Sigma$, so it is in $\Sigma$. 
**Note that we have used the fact that a finite intersection of sets in $\Sigma$ is also in $\Sigma$.  
