Here's my answer.Let $\displaystyle f:[-1,1] \rightarrow \mathbb{R}$ be an odd regulated function. Prove that:

$\displaystyle \int_{-1}^1f=0$.

By definition of a regulated function $\displaystyle \int_{-1}^1f= \lim_{n \rightarrow \infty} \phi_n$ where $\displaystyle \phi_n$ is a sequence of step functions converging uniformly to f.

Claim: $\displaystyle \phi_n$ is odd.

We already know that $\displaystyle \lim_{n \rightarrow \infty} \phi_n(x)=f(x)$. (*)

We can multiply both sides by -1 to get $\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)$ for some $\displaystyle x \in [-1,1]$.

Since (*) is valid $\displaystyle \forall \ x \in [-1,1]$, we also know that:

$\displaystyle \lim_{n \rightarrow \infty}\phi_n(-x)=f(-x)$.

But we can rewrite the first one as $\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)=f(-x)$ since we know that f is odd.

We can now equate the two to get:

$\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=\lim_{n \rightarrow \infty} \phi_n(-x)$

So we know that $\displaystyle \phi_n$ is odd.

My next claim is:

$\displaystyle \phi_k:[-1,1] \rightarrow \mathbb{R}$ is an odd step function so $\displaystyle \int_{-1}^1 \phi_k=0$ where $\displaystyle k=0,1, \ldots$.

$\displaystyle \int_{-1}^1 \phi_k==\int_{-1}^0 \phi_k+\int_0^1 \phi_k=\int_{-1}^0 \phi_k(x)~dx+\int_0^1 \phi_k(x)~dx$.

We can use the fact $\displaystyle \phi_k$ is odd:

$\displaystyle =\int_{-1}^0 \phi_k(x)~dx-\int_0^1 - \phi_k(x)~dx=\int_{-1}^0 \phi_k(x)~dx-\int_0^1 -\phi_k(-x)~dx$

We can then rewrite this as:

$\displaystyle =\int_{-1}^0 \phi_k(x)~dx-\int_{-1}^0 \phi_k(x)~dx=0$

Is this the right?