Show that f is measurable. Let $a > 0, b \geq 0$ and the function $f: \mathbb{R} \to \mathbb{R}$
$$f(x) = \left\{\begin{matrix}
1, & |x| \leq a \\ 
 b & |x| > a
\end{matrix}\right.$$
show that it is measurable.
I don't have clue about to show this!!! 
 A: I assume measurability with respect to the Borel $\sigma$-algebra.
You can rewrite $f$ as:
$$f(x)=\mathbf{1}_{[-a,a]}(x)+b\mathbf{1}_{\mathbb{R}\setminus [-a,a]}(x)$$
$[-a,a]$ is an interval so it is measurable. Thus $\mathbb{R}\setminus [-a,a]$ is measurable too. This simplies that the indicator functions $\mathbf{1}_{[-a,a]}$ and $\mathbf{1}_{\mathbb{R}\setminus [-a,a]}$ are measurable functions. You can then conclude by using the fact that the sum and product of measurable functions is still measurable.
A: Hint: Characteristic functions of measurable sets are measurable, so is any linear combination of them. Also when it is not specified euclidean spaces are considered to be equipped with the Borel sigma algebra, which is generated by the collection of open sets.
A: I don't think it's necessarily measurable. Did you mean Lebesgue measurable? If so, then I think so.

Let $a, b \in \mathbb R$.
Define $A := \{x | |x| \le a\}$.
Consider a measure space $(S, \Sigma, \mu)$.
$1$ is constant and hence $\Sigma$-measurable. $(**)$
$b$ is constant and hence $\Sigma$-measurable.
If $A \in \Sigma$, then:


*

*$1_A$ is $\Sigma$-measurable. $(**)$

*$A^C \in \Sigma$

*$1_{A^C}$ is $\Sigma$-measurable.
If $A^C \in \Sigma$, then:


*

*$1_{A^C}$ is $\Sigma$-measurable.

*$A \in \Sigma$

*$1_{A}$ is $\Sigma$-measurable.
So, if $A \in \Sigma$, then
$$f = 1 \times 1_A + b \times 1_{A^C}$$
since finite sums of $\Sigma-$measurable functions are $\Sigma-$measurable and finite products of $\Sigma-$measurable functions are $\Sigma-$measurable.
If our measure space is $(\mathbb R, \mathscr B(\mathbb R), \lambda)$, then $A \in \mathscr B(\mathbb R)$ because $A$ is an interval.

$(**)$ Note: We say a function $f$ is $\Sigma-$measurable from $(S, \Sigma, \mu)$ to $(\mathbb R, \mathscr B(\mathbb R))$ if
$\sigma(f) := \{f^{-1}(B) | B \in \mathscr B(\mathbb R)\} \subseteq \Sigma$
Now, if $f$ is a constant, then
$\sigma(f) = \{\emptyset, S\} \subseteq \Sigma$
If $f = 1_B$ for $B \in \Sigma$, then
$\sigma(f) = \sigma(B) \subseteq \Sigma \tag{*}$
$(*)$ Note: $B \in \Sigma \to \sigma(B) := \{\emptyset, B, B^C, S\} \subseteq \Sigma$
