TITLE: Showing uniqueness of Riemann's Integral
QUESTION [5 upvotes]: I am given the definition: Le $f$ be defined on $[a,b]$. we say that $f$ is Riemann Integrable on $[a,b]$ if there is a number $L$ with the following property: for every $\epsilon>0$, there is a $\delta > 0$ such that $\left\|P\right\|< \delta$ implies $| \sigma -L| < \epsilon$ where $\sigma$ is the Riemann Sum of $f$ over the partition $P$ of $[a,b]$. In this case, we say that $L$ is the Riemann Integral of $f$ over $[a,b]$, and write $\int_{a}^{b} f(x) dx=L$
I am then asked to show why $L$ is a unique limit. It does make sence if not unique but how to show is well above me.

REPLY [7 votes]: Assume that $L_1$ and $L_2$ are the Riemann inttegrals of $f$ over $[a,b]$. We want to show that $L_1=L_2$. Let $\epsilon >0$. Then for each $i=1,2$, there exists $\delta_i>0$ such that
$$\|P \|<\delta_i \quad \Rightarrow \quad |\sigma-L_i|<\frac{\epsilon}{2}$$
whenever $P$ is a partition of $[a,b].$ Take $\delta$=min $\{\delta_1,\delta_2\}$. Fix a partition $P$ of $[a,b]$ and suppose $\|P\|<\delta$. Note that $\delta\le \delta_i$ for $i=1,2.$ Hence
$$0\le|L_1-L_2|\le|\sigma-L_1|+|\sigma-L_2|<\epsilon.$$
Since $\epsilon>0$ was arbitrary, 
$$0\le|L_1-L_2|<\epsilon$$
holds for all $\epsilon >0.$ This forces us to conclude that $|L_1-L_2|=0.$ Hence, $L_1=L_2.$