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Hello I am wanting to know if this proof makes sense.

If $f$ is Riemann integrable on $[a,b]$ and $f$ is bounded on this interval by some $M \in \mathbb{R}$, show that $$\left\lvert \int_{a}^{b} f \right\rvert \le M(b-a).$$

I think it could also be proved using basic properties such as that if $f(x) \le g(x)$ then so, too, are the integrals, but I am especially wondering about the following.

By definitions we would have all $\epsilon \gt 0$ , $\exists \delta \gt 0$ such that for all partitions of mesh less than $\delta$

$$0 \le \left\lvert \sum_{i=1}^{n} f(t_{i})(x_{i}-x_{i-1})-L \right\rvert \lt \epsilon$$

but since $f$ is bounded,

$$0 \le \left\lvert M\sum_{i=1}^{n}(x_{i}-x_{i-1})-L \right\rvert.$$

And

$$0 \le | M(b-a)-L |$$ since it must hold for any such partitions and tags.

Is this valid, does it make sense?

Thank you all

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  • $\begingroup$ with non-regular mesh it is the Stejes integral not the Riemann integral, so set $x_i = a + i/n$ $\endgroup$
    – reuns
    Jan 23, 2016 at 21:40

2 Answers 2

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You correctly argued that for sufficiently fine partitions we have

$$L - \epsilon \leqslant \sum_{k=1}^n f(t_k)(x_k - x_{k-1}) \leqslant L + \epsilon.$$

The ensuing steps, however, do not lead to the desired conclusion. For example, $0 \leqslant |M(b-a) - L|$ does not guarantee that $|L| < M(b-a).$

To complete the proof, argue as follows.

If $L > M(b-a)$ choose $\epsilon$ such that $L - \epsilon > M(b-a)$ and you find sums greater than $M(b-a)$, a contradiction. Make a similar argument to show we cannot have $L < -M(b-a)$.

Another approach is:

If $f$ is bounded, then $|f(x)| \leqslant M$ for $x \in [a,b],$ and

$$-M \leqslant f(x) \leqslant M.$$

Let $P = (x_0,x_1, \ldots, x_n)$ be any partition of $[a,b].$

Then for all $k = 1, 2, \ldots, n$ we have

$$-M \leqslant \inf_{x \in [x_{k-1},x_k]}f(x) \leqslant \sup_{x \in [x_{k-1},x_k]}f(x)\leqslant M.$$

Hence, for any lower and upper sum

$$-M(b-a) \leqslant L(P,f) \leqslant \int_a^b f(x) \, dx \leqslant U(P,f) \leqslant M(b-a),$$ and

$$\left|\int_a^b f(x) \, dx\right| \leqslant M(b-a). $$

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  • $\begingroup$ How would I go about choosing such $\epsilon$ s to make $L- \epsilon \gt M(b-a)$ and making a contradiction? Could you elaborate on that? $\endgroup$
    – Quality
    Jan 24, 2016 at 23:46
  • $\begingroup$ Sure. Assume $M(b-a) < L$. Choose any positive number $\epsilon$ such that $\epsilon < L - M(b-a)$. Then $M(b-a) < L - \epsilon < \sum f(t_i)(x_i - x_{i-1})$ for fine enough partitions. This is a contradiction since any sum must be less than or equal to $M(b-a)$. $\endgroup$
    – RRL
    Jan 25, 2016 at 0:11
  • $\begingroup$ Alright, and if I include that, then is my proof valid? $\endgroup$
    – Quality
    Jan 25, 2016 at 0:15
  • $\begingroup$ Yes -- that shows that $L < M(b-a)$. A similar argument shows $-M(b-a) < L$ and finally $|L| < M(b-a)$. $\endgroup$
    – RRL
    Jan 25, 2016 at 0:17
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My version in terms of the limit of sums interpretation of the Riemann integral:

Define $x_i=a+i/n$. As $f$ is bounded by $M$ on the interval, for all $n \in \mathbb{N}$

$$|\sum_{i=1}^{n}f(x^{*}_{i})[x_i-x_{i-1}]| \leq \sum_{i=1}^{n}|f(x^{*}_{i})[x_i-x_{i-1}]|\leq \sum_{i=1}^{n}|M[x_i-x_{i-1}]|$$

$$|\sum_{i=1}^{n}f(x^{*}_{i})[x_i-x_{i-1}]| \leq M(b-a) \,...(1)$$

Now as $f$ is Riemann integrable on $[a,b]$ then

$$\lim_{n \to \infty} \sum_{i=1}^{n}f(x^{*}_{i})[x_i-x_{i-1}]=\int_{a}^{b}f(x)dx$$

So by $(1)$ then we have

$$|\int_{a}^{b}f(x)dx|\leq\lim_{n \to \infty}M(b-a)=M(b-a) $$

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