I'm hoping someone could look over my proof attempt of the following claim.
The Statement
For the step function $\phi$ on the compatible partition $P=\{p_0,...p_k\}$. Then we say that the function $I:[a,b]\rightarrow$R with $I(t)=\int_a^t\phi(x)$$dx$ has the following properties:
(a) $I(t)$ is continuous on $[a,b]$.
(b) $I$ is differentiable on the union over the partitions $(p_{i-1},p_i)$ and$I'(t)=\phi (t)$.
Proof Attempt
I have attempted to prove the (a) and (b) by considering a refinement of the partition such that we define the new partition $$\tilde{P}=\{p_0,...(p_j=t) ...p_k\}$$ If we now evaluate $I$ this gives us the result:
$$I(t)=\int_a^t\phi(x) dx=\sum_{i=1}^j\phi_i(p_i-p_{i-1})$$
Note: $\phi_i$ denotes the constant value that the step function takes on the interval $(p_{i-1},p_i)$.
As this is the finite sum of real valued constants, this implies that $I(t)$ is continuous and the derivative exists (being equal to $\phi (t))$.
The claim feels fairly simple, but I'm having trouble showing this more rigorously.