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.
Step One
First define the new partition on $[a,t]$ which we call $\tilde{P}=\{(a=p_0),p_1,$ [$...$] $p_{n-1}, (\tilde{p}_n=t)\}$.
Step Two
If we now evaluate $I$ this gives us the result:
$$I(t)=\int_a^t\phi(x) dx=\sum_{i=1}^j\phi(i)(p_i-p_{i-1})=t\phi(t)+ C $$
(where C denotes a constant formed by the other terms in the sum).
Step Three
We observe that $I(t)$ is a linear function in $t$ as it is in the form $I(t)= at + C$ where $a=\phi (t)$. And so it is clearly continuous on the $[a,b]$ since it is continuous for all $t\in [a,b]$.
We also note that as it is linear, the derivative is simply $I'(t)=\phi (t)$. Again, since this holds for $t\in [a,b]$, then we are able to conclude that this holds over the each subinterval of $P$.