Let $V$ be the set consisting of all step functions $[0,1]→\mathbb{R}$.
(a) Prove that $V$ is a vector space.
(b) Prove that the indefinite integral gives a linear map $V→\mathcal{F}([a, b], \mathbb{R})$.
(c) Is this map is injective? Why or why not?
(a) We know that $\mathcal{F}([0,1], \mathbb{R})$ is a vector space, so it suffices to show that the set of step function is a subspace. But we know that if $s$ and $t$ are step functions, then so is $s+t$ (by passing to a common refinement). And if $s$ is a step function and $c \in \mathbb{R}$, then $c s$ is a step function (by using the same refinement as for $s$ ). Hence the set of step functions is closed under addition and scalar multiplication, so it is a subspace.
(b) This is an immediate consequence of linearity (and homogeneity) for integrals of step functions: $\int_0^x(s+t)(y) d y=\int_0^x s(y) d y+\int_0^x t(y) d y$ and $\int_0^x c s(y) d y=c \int_0^x s(y) d y$. So if we let $\phi(s)=\int_0^x s(y) d y$, then $\phi(s+t)=\phi(s)+\phi(t)$ and $\phi(c s)=c \phi(s)$.
(c) This is not injective: both $s(x)=0$ and
\[
t(x)=\left\{\begin{array}{l}
1 \text { if } x=0 \\
0 \text { if } x \neq 0
\end{array}\right.
\]
have $\phi(s)=\phi(t)=0$