For a convergence theorem on integral see: Riemann Integral: Uniform Convergence

For an improper version of integral see: Riemann Integral: Improper Version

For a comparison of integrals see: Uniform Integral vs. Riemann Integral


Given a finite measure space $\mu(\Omega)<\infty$ and a Banach space $E$.

Consider functions $F:\Omega\to E$.

Define the generalized Riemann integral by: $$\int F\mathrm{d}\mu:=\lim_\mathcal{P}\{\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)\}_\mathcal{P}$$ over finite measurable partitions: $$\mathcal{P}\subseteq\Sigma:\quad\Omega=\bigsqcup_{A\in\mathcal{P}}A\quad(\#\mathcal{P}<\infty)$$ being ordered by refinement: $$\mathcal{P}\leq\mathcal{P}':\iff\forall A'\in\mathcal{P}'\exists A\in\mathcal{P}:\quad A\supseteq A'$$ (In fact, the tags are just surpressed.)


What do nonintegrable functions look like? $$F:\Omega\to E:\quad F\notin\mathcal{L}_\mathfrak{R}(\mu)$$ Obviously, nonmeasurable nonexamples are countless: $$\mu^*(V)\neq\mu_*(V):\quad F=\chi_V$$ Also, a.e. boundedness is a necessary condition on integrability: $$F\notin\mathcal{L}^\infty(\mu)\implies F\notin\mathcal{L}_\mathfrak{R}(\mu)$$ So are there bounded measurable nonexamples?
(Note that continuity is not an accessible concept here.)

Besides, pathological examples like the Dirichlet function become integrable.

Deep Consequence

Riemann integrability is not guaranted by absolute integrabilit alone: $$\int\|F\|\mathrm{d}\mu<\infty\nRightarrow \int F\mathrm{d}\mu\in E$$ But this shouldn't be to big surprise as this wasn't either the case for Bochner integrability.

  • $\begingroup$ Probably any nonmeasurable function (modulo equality a.e. or w.r.t. the completion of the measure space) will not be "Riemann"-integrable. BTW: I am still thinking about your related question. $\endgroup$ – PhoemueX Nov 14 '14 at 10:44
  • $\begingroup$ Oh nice. :) So it's still missing wether uniform implies Riemann. $\endgroup$ – C-Star-W-Star Nov 14 '14 at 13:40
  • $\begingroup$ Yeah thought so too like $\chi_A$ for $A\notin\hat{\Sigma}$. I will add the assumption of measurability. So but how about measurables? $\endgroup$ – C-Star-W-Star Nov 14 '14 at 13:42
  • $\begingroup$ @PhoemueX: If you're still interested, see below. ;) Banach-space valued function: Bounded and measurable (though separable valued nor totally bounded) but not Riemann integrable. $\endgroup$ – C-Star-W-Star Nov 25 '14 at 16:41

Ok, I think I got it now...

Take a slight variation of the famous example: $$F:[0,1]\to\ell[0,1]:t\mapsto\chi_t$$

At least, it is bounded: $\|F(t)\|\equiv1$.

Especially, it is absolutely integrable: $\int\|F(t)\|\mathrm{d}t=1$

However, it is not measurable as: $$\|\chi_s-\chi_t\|^2=2\quad(s\neq t)$$ so taking a Vitali set yields: $$U:=\bigcup_{v\in V}B_{\frac{1}{\sqrt{2}}}(\chi_v):\quad U\in\mathcal{T}[0,1],F^{-1}U=V\notin\mathcal{B}[0,1]\quad\left(V\subseteq[0,1]\right)$$

Moreover, it has nonseparable range: $$\|\chi_s-\chi_t\|\equiv2\quad(s\neq t)$$ So it can't be pointwise limit in any case: $S_n\nrightarrow F$

Finally, it is not Riemann integrable as: $$\|\sum_{A\in\mathcal{P}}F(a)\mu(A)-\sum_{A\in\mathcal{P}}F(a')\mu(A)\|=\sum_{A\in\mathcal{P}}\|\chi_a-\chi_{a'}\|\mu(A)=2\sum_{A\in\mathcal{P}}\mu(A)\equiv2\mu(\Omega)\quad(a\neq a')$$


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