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Prove if $\int_{a}^{b} f(x)$ exists, then for every $\epsilon>0$, there is a $\delta >0 $ such that $|\sigma_1 -\sigma_2|<\epsilon$, if $\sigma_1$ and $\sigma_2$ are Reimann sums of $f$ over partitions P$_1$ and P$_2$ of $[a,b] $ with norms less than $\delta$.

First Recalling the primary definition: As $\int_{a}^{b} f(x)$ exists, there is a unique $L$ s.t. $L = \int_{a}^{b} f(x)$ and for every $\epsilon>0$, there is a $\delta >0 $ s.t.

$$|\sigma – L| < \epsilon$$

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Second, I want to show that because $f$ is integrable on $[a,b]$, the function has to be bounded on $[a,b]$:

If $f$ is unbounded on $[a,b]$, for any partition P$_1$,P$_2$$\in P$

we have with $M>0$

$$|\sigma_1 – \sigma_2|>M$$

Third, Because $\int_{a}^{b} f(x)$ exists $f$ has to be bounded

I am new to this. I have to admit, I am quite lost. I just do not see how i can prove part 2 . Any help is much appreciated.

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Fix $\epsilon>0$ and let $L = \int_a^b f(x) \ dx$.

The normal definition states that there is a $\delta>0$ such that if $\sigma$ is a Riemann sum such that its partition $P_{\sigma}$ satisfies $\text{mesh}\left(P_{\sigma}\right) < \delta$, then

$$|\sigma – L| < \epsilon$$

Now consider $\frac{\epsilon}{2}$. The above definition implies there is a $\delta_1>0$ such that if $\sigma_1, \sigma_2$ are arbitrary Riemann sums such that $\max\left\{\text{mesh}\left(P_{\sigma_1}\right), \text{mesh}\left(P_{\sigma_2}\right)\right\} < \delta_1$, then we have

$$|\sigma_1 – L| < \frac{\epsilon}{2}$$

and

$$|\sigma_2 – L| < \frac{\epsilon}{2}$$

and hence, by the triangle inequality,

$$|\sigma_2 – L| + |\sigma_1 – L| = |L – \sigma_2| + |\sigma_1 – L| \leq |\sigma_1 – \sigma_2| < \epsilon$$

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