Suppose that $s$ is a step function, and $f(x)=mx+k$. How to prove that $s(f(x))$ is step function also?
Edit:
Definition of step function:
Function $s$ defined on $[a,b]$ will called a step function is there exist a partition $P$ of $[a,b]$, $P=\{a=x_0,...,x_n=b\}$ such that $\displaystyle s_{|{\large{(x_{i-1},x_i)}}}=t_i$, for some $t_i\in \Bbb R$.
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$\begingroup$Hints: The step functions are finite linear combinations of $$s_a\colon x\mapsto\begin{cases}0 & x<a \\ 1 & x\geq a\end{cases}$$ $$\hat{s}_a\colon x\mapsto\begin{cases}0 & x\leq a \\ 1 & x> a\end{cases}$$ $$c\colon x\mapsto 1$$ Now the only thing you have to show is that $s_a\circ f$, $\hat{s}_a\circ f$ and $c\circ f$ are step functions. In proving this you might want to consider $m<0$, $m=0$ and $m>0$ separately.
Exempli gratia: Let $m>0$, then $$s_a\circ f\colon x\mapsto s_a(mx+k) = \begin{cases}0 & mx+k<a \\ 1 & mx+k\geq a \end{cases} = \begin{cases}0 & mx<a-k \\ 1 & mx\geq a-k \end{cases}$$ Since $m>0$, $mx<y$ if and only if $x<\frac{y}{m}$, thus $$s_a\circ f\colon x\mapsto \begin{cases}0 & x<\frac{a-k}{m} \\ 1 & x\geq \frac{a-k}{m} \end{cases} = s_{\frac{a-k}{m}}(x)$$ and we conclude $s_a\circ f = s_{\frac{a-k}{m}}$.
The other cases are mostly similar.
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