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Lebesgue's criterion for Riemann-integrability says that a function $f:[a,b]\to\mathbb{R}$ is Riemann-integrable iff it is bounded and the set of points at which it is not continuous has measure zero.

This can be easily extended to functions with values in $\mathbb{R}^n$. However, is there an equivalent criterion for functions taking values in a Banach space or, making some more assumptions, in a (separable) Hilbert space?

To avoid confusion: The Riemann integral for Banach space valued functions is (obviously) not defined via upper and lower sums, but as described in for real-valued functions

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Have you tried adapting the proof for $\mathbb R$ to see what happens? – GEdgar Jun 26 '12 at 16:24
The proof heavily uses the fact that the reals are ordered, in particular using upper and lower Riemann sums and I thus don't see how to adapt it. – ChrisBln Jun 26 '12 at 16:26
First prove your result when $f$ is continuous. Then prove it when $f$ is continuous except at one point. By then you should be able to see what to do in general. – GEdgar Jun 26 '12 at 16:27
If $f$ is continous, then it is integrable. If it continuous except at finitely many points, I can always chose partitions that take points very closely around the points of discontinuity so that they "don't matter". However, I don't see how to extend this to any set of measure zero. – ChrisBln Jun 26 '12 at 16:31
First: you said "continuous therefore integrable". Review that proof. Next: review the real-valued proof, and see what they do with the "points of discontinuity". – GEdgar Jun 26 '12 at 16:40
up vote 8 down vote accepted

Let $\mathbf E$ be a Banach space, let $a<b$ be real numbers, let $f : [a,b] \to \mathbf E$ be a function.

A partition $\pi$ of $[a,b]$ is a finite subset, $\{a,b\} \subseteq \pi \subset [a,b]$, usually written in order: $a = x_0 < x_1 < \dots < x_n = b$. Tags for a partition $\pi$ as above are points $t_i$ such that $x_{i-1} \le t_i \le x_i$ for $1 \le i \le n$. Partition $\pi_1$ refines partition $\pi_2$ iff $\pi_1 \supseteq \pi_2$, remembering that a partition is a finite set.

Definition. Let $f$ be as above, and let $\mathbf u \in \mathbf E$. We say that $f$ is Riemann integrable on $[a,b]$ and $\mathbf{u}$ is its integral iff: for every $\epsilon > 0$, there is a partition $\pi_0$ of $[a,b]$ such that for all refinements $\pi=(x_i)_{i=0}^n$ of $\pi_0$ and all tags $(t_i)_{i=1}^n$ for $\pi$, $$ \left\|\mathbf u - \sum_{i=1}^n f(t_i)\;(x_{i}-x_{i-1})\right\| < \epsilon. $$

Lemma $f$ is integrable iff: for every $\epsilon > 0$ there is a partition $\pi = (x_i)_{i=0}^n$ such that for any two choices $(t_i)_{i=1}^n, (s_i)_{i=1}^n$ of tags for $\pi$, we have $$ \left\|\sum_{i=1}^n \big(f(t_i)-f(s_i)\big)\;(x_{i}-x_{i-1})\right\| < \epsilon. $$

Proof. Cauchy criterion.

Theorem. Let $f : [a,b] \to \mathbf E$ be bounded and continuous except on a set $N\subseteq [a,b]$ of measure zero. Then $f$ is integrable.

Proof. Add $\{a,b\}$ to the null set $N$ to avoid special cases for endpoints. Let $\epsilon>0$. Say $f$ is bounded by $M$, $\|f(x)\| \le M$. Let $\alpha > 0$ be so small that $2M\alpha + \alpha(b-a) < \epsilon$. For an open interval $(u,v)$ we say $f$ has oscillation at most $\alpha$ on $(u,v)$ if for all $x,y \in (u,v)$, $\|f(x)-f(y)\| \le \alpha$. If $f$ is continuous at a point $s$, then there is an inverval $(u,v)$ with rational endpoints, $s \in (u,v)$, so that $f$ has oscillation at most $\alpha$ on $(u,v)$. So there is a countable union of such intervals $(u,v)$ that contains $[a,b] \setminus N$, and thus has full measure. So there is a finite list $(u_j,v_j)$ of intervals where $f$ has oscillation at most $\alpha$, and their union has measure greater than $b-a-\alpha$. Then there is a partition $\pi = (x_i)_{i=0}^n$ of $[a,b]$ such that each subinterval $[x_{i-1},x_i]$ from the partition either is contained in an interval where $f$ has oscillation at most $\alpha$, or is an "exceptional" interval. The total length of all the exceptional intervals is ${}< \alpha$. Now let $(t_i)$ and $(s_i)$ be two choices of tags for the partition $\pi$. Now we must consider $$ \left\|\sum_{i=1}^n\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\| \le \sum_{i=1}^n\left\|\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\| . $$ Consider the term $(f(t_i)-f(s_i))(x_i-x_{i-1})$. If the subinterval $[x_{i-1},x_i]$ is not exceptional, then $$ \left\|\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\| \le \alpha (x_{i}-x_{i-1}) , $$ so the total of all terms for non-exceptional intervals is at most $\alpha (b-a)$. If the subinterval $[x_{i-1},x_i]$ is exceptional, then $$ \left\|\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\| \le 2M (x_{i}-x_{i-1}) , $$ so the total of all terms for exceptional intervals is at most $2M\alpha$. Thus $$ \left\|\sum_{i=1}^n\big(f(t_i)-f(s_i)\big)(x_i-x_{i-1})\right\| \le \alpha(b-a)+2M\alpha < \epsilon . $$

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Great. I was actually not at all convinced the criterion would hold in this case. Thanks a lot. Should have had the idea to replace the upper and lower sum by (some kind of) Cauchy criterion myself, though... – ChrisBln Jun 26 '12 at 18:48

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