I've attempted to tackle the following problem and I'm not sure about my solution:
$f:\:\left[-1,1\right]\:\rightarrow \:\mathbb{R}$ is an odd and integrable function. Show that: $$\int _{-1}^1\:\:f\left(x\right)dx=0$$
This can be done by showing $\int _0^1\:\:f\left(x\right)dx=-\int _{-1}^0f\left(x\right)dx\:$. My approach was thus:
If $P$ is a partition of $\left[0,1\right]$ containing $n+1$ elements, then $-P=\left\{-p\::\:p\in P\right\}$ is evidently a partition of $\left[-1,0\right]$ with the same number of elements and symmetrical intervals. Likewise, for the set $T$ which contains an arbitrary point for each "subinterval" of $P$, the set $-T=\left\{-t\::\:t\in T\right\}$ contains a symmetrical arbitrary point for each subinterval of $-P$.
Therefore: $$\int _0^1f\left(x\right)dx=\lim _{λ\left(P\right)\to 0}\left(R\left(f,P,T\right)\right)\:\:=\:\lim \:_{λ\left(P\right)\to \:0}\left(\sum _{i=1}^n\left(Δx_i\cdot f\left(t_i\right)\right)\right)\: =_{oddity\:of\:the\:function}\:\lim \:\:_{λ\left(P\right)\to \:\:0}\left(\sum \:_{i=1}^n\left(Δx_i\cdot \:-f\left(-t_i\right)\right)\right)\: =\:-\lim \:\:_{λ\left(P\right)\to \:\:0}\left(\sum \:_{i=1}^n\left(Δx_i\cdot \:f\left(-t_i\right)\right)\right)=-\lim \:_{λ\left(-P\right)\to \:0}\left(R\left(f,-P,-T\right)\right)=-\int _{-1}^0f\left(x\right)dx\:\:$$
Is that a sufficient proof? I'm afraid it's an abuse of notation and I'm not allowed to manipulate the limit of the Riemann sum so cavalierly.