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. $$|\sigma - L| < \epsilon$$

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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    $\begingroup$ Are you trying to prove the second part that $f$ is bounded in addition to the first proposition? $\endgroup$ – RRL Jul 26 '17 at 3:50
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    $\begingroup$ See here $\endgroup$ – RRL Jul 26 '17 at 3:50
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    $\begingroup$ This is a relatively simple problem as one answer here indicates. The converse is slightly tricky to prove. It is the usual Cauchy's principle of convergence for limit of Riemann sums. $\endgroup$ – Paramanand Singh Jul 26 '17 at 8:12
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    $\begingroup$ Definition of Riemann integral assumes that the function is bounded. This is because the way limit of Riemann sums are defined, the existence of limit ensures that the function has to be bounded. The linked answer by @RRL gives more details on this. $\endgroup$ – Paramanand Singh Jul 26 '17 at 8:15
  • $\begingroup$ I thought I had to prove that $f$ is bounded first. confusion from my part. $\endgroup$ – gegu Jul 26 '17 at 11:55

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

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


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

and hence, by the triangle inequality,

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

  • $\begingroup$ thx for the input. Just trying to understand: isnt the triangle inequality stating the opposite of what you wrote: such as for example $|x-y| \leq |x-w|+|w-y|$. ? $\endgroup$ – gegu Jul 26 '17 at 12:33
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    $\begingroup$ @gegu Oh, welp, that was a stupid mistake. It doesn't affect the argument though. Just start the inequality with $|\sigma_{1} - \sigma_{2}|$ instead. $\endgroup$ – MathematicsStudent1122 Jul 26 '17 at 14:22

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