How does one show if one function $u:[a,b]\to \mathbb{R}$ is measurable?
Using Schilling's methodology, I understand this as to show that $u$ is $\mathcal{A}/\mathcal{B}$ measurable (see Chapter 8), where $\mathcal{A}$ is a $\sigma$-algebra on $[a,b]$. To do this, one may show that, for example, $\{u<\alpha\}\in \mathcal{A}$ for all $\alpha\in \mathbb{R}$.
Is this automatically true, if $u$ were continuous on $[a,b]$?
Here is what I did,
$\{u<\alpha\}=\{x\in [a,b]:u(x)<\alpha\}=u^{-1}((-\infty,\alpha))$
Then the right hand side is open, because of continuity. But does it belong to $\mathcal{A}$?
Another question is, no sure if it is relevant, what if $u$ were not continuous on $[a,b]$?
Consider 2 examples:
(1) $\mathcal{A}=\sigma\{[a,(a+b)/2),[(a+b)/2,b]\}$ and $u(x)=x$ ($u$ is not $(\mathcal{A},\mathcal{B}_{\mathbb{R}})$-measurable).
(2) $\mathcal{A}$ is the Borel $\sigma$-algebra (i.e. generated from open sets) and $u$ is continuous ($u$ is $(\mathcal{A},\mathcal{B}_{\mathbb{R}})$-measurable according to your argument).
Regarding your second question, there are discontinuous functions that are measurable w.r.t. $\mathcal{A}$ in (1), e.g. $u(x)=1_{[a,(a+b)/2)}(x)$, and there exist functions that are not measurable w.r.t. $\mathcal{A}$ in (2) (which is much larger), e.g. $u(x)=1_N(x)$, where $N$ is a non-Borel set in $[a,b]$.