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Exercise: Show that if $f: [a,b] \to X$,a complete normed vector space, is Riemann integrable, then $f$ is bounded.

Definition of Riemann-integrable:

Let $f: [a,b] \to X$ be a map. F is Riemann-integrable if for every $\epsilon >0$ there exists a step function $u: [a,b]\to X$ and a step-function $v:[a,b]\to \mathbb{R}$ such that:

$(\forall x\in[a,b]: \| f(x)-u(x)\|\leq v(x))$ and $\mathbb{I}v<\epsilon$

I just don't see where to start with this one.

Am I supposed to (1) argue contra-positively or (2) show that if $\forall x\in[a,b]:\| f(x) - u(x)\|\leq v(x)$ and $v(x)$ can become arbitrarily small for $\forall x\in [a,b]$, then that must mean that $u(x)$ at some $x$ achieves the max value $M$, ($|f(x)| \leq M$) , that bounds $f(x)$?

Mathaniel
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2 Answers2

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If $f: [a,b] \to \mathbb{R}$ is Riemann Integrable then we have that $\inf_{P}U(f,P)=\sup_{P}L(f,P)$ where these are the lower and upper Darboux Sums, and P is a partition of $[a,b]$. For any particular partion of [a,b] we have that $U(f,P)=\sum\limits_{k=1}^{n}{(x_k-x_{k-1})M_k}$ where $M_k=\sup_{x_{k-1} \leq x \leq x_k}{f(x)}$. Suppose for sake of contradiction that f(x) is unbounded on [a,b] This implies that $\forall M > 0$ there exists $x \in [a,b]$ such that $f(x) > M/||P||$ where $||P||$ is the width of the partition.

Then consider for any partition P the Darboux sum will be bounded below by $||P||M/||P||=M \to \infty$. Hence, the Darboux sum will diverge and thus the function cannot be Riemann Integrable.

ADA
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  • what is weird (and I dont see clearly the reason) is that some improper (unbounded) Riemann integrals are convergent... by example $$\int_0^1\frac{\mathrm dx}{\sqrt x}=2$$ –  Apr 22 '17 at 19:13
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    Yes this function is improperly riemann integrable. So what does that mean, we can cut out a neighborhood about $x=0$ where the function is bounded and this will be riemann integrable. To be improperly Riemann integrable means that as we take the limit with $x \to 0^{+}$ the integral will be finite.

    It is interesting, and you attempt to construct a partition of the interval $[0,1]$ where the Darboux sum of $1/\sqrt{x}$ is divergent.

     – ADA Apr 22 '17 at 19:19
  • In the process of arguing by contradiction, @ADA 's post missed the case that $f$ may be unbounded from below. The argument for this case need to be modified a bit. Now suppose $f$ is unbounded from below. Let $I_k :=[x_{k-1},x_k]$. Let $A_f :={i\in {1,2, ..., n} \vert f\text{ is unbounded from below on }I_i}.$ By our assumption, $A_f$ is nonempty. Then dividing the summation into two parts, i.e. one over $A_f$ and the other one over ${1, 2, ..., n}\smallsetminus A_f.$ The rest is the same as in the case that $f$ is unbounded from above. – Sam Wong Feb 27 '20 at 15:45
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Take $\epsilon=1.$ Then there exist step functions $u,v$ such that

$$\|f(x)-u(x)\| \le v(x) \le 1,\,\, \text { for all } x\in [a,b].$$

Thus

$$\|f(x)\| \le \|f(x)-u(x)\| + \|u(x)\| \le 1 + \|u(x)\|\,\, \text { for all } x\in [a,b].$$

Since a step function is bounded, $\|u\|$ is bounded, hence so is $1 + \|u\|,$ hence so is $f.$

zhw.
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