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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.

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2 Answers 2

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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, $L_1=L_2.$

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    $\begingroup$ I know this is an old post, but I don't follow your second last sentence. How does it force us to say they're equal to $0$? What if it's greater than $0$? $\endgroup$ Dec 14, 2017 at 21:09
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    $\begingroup$ @ContraModernistae For each $n\geq 1$, we have $0\leq |L_1-L_2|<\frac{1}{n}$, then apply the Squeeze Theorem $\endgroup$
    – Juniven
    Dec 14, 2017 at 22:46
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    $\begingroup$ @ContraModernistae Generally, if $a$ and $b$ are real numbers and $a<b+\epsilon$ for all $\epsilon>0$, then $a\leq b$. Can you prove this? Now, in the proof that I have presented, I said that $|L_1-L_2|<0+\epsilon$ for all $\epsilon>0$. With this we conclude that $|L_1-L_2|\leq 0$. But we know that $|L_1-L_2|\geq 0$, and hence equality follows. $\endgroup$
    – Juniven
    Dec 14, 2017 at 22:52
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suppose there is another limit different form $L$ call it $P$ then $\exists \epsilon : |L-P|>2\epsilon$ now by definition $2\epsilon>|\sigma-L|+|\sigma-P|\geq|L-P|>2\epsilon$

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