how to find a Borel function $f$ such that $a(t) = f(b(t))$ Let $b(t)$ be a non-increasing, right-continuous and positive function on $(0,\infty)$ (clearly, it is Borel) and $a(t)$ be a measurable function on $(0,\infty)$. Moreover, $a(t)$ is a constant on an interval when $b(t)$ is a constant on this interval.
Can we find a Borel function $f:\sigma(b(t))\rightarrow R$ such that $a(t) = f(b(t))$?

In general, no: if $L$ is the Cantor function, let $b=2-L$ and $a=2+1_{S}$, where $1_S$ is the indicator function of a non-Borel subset $S$ of the Cantor set $C$. Then $a$ is non-Borel and constant on all the intervals where $b$ is constant (i.e. on $\Bbb R\setminus C$), but the RHS you require is clearly a Borel function.
