# Bochner Integral vs. Riemann Integral

Disclaimer

Anyway, it is written as problem. Have fun! :)

Reference

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Prologue

This is the final comparison.

All notions are up to null sets.

Definitions

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

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

Denote the Bochner measurable functions by $\mathcal{B}$.

Define the Bochner integral by: $$F\in\mathcal{B}:\quad\int F\mathrm{d}\mu:=\lim_n\int S_n\mathrm{d}\mu$$ for simple functions satisfying: $$\int\|F-S_n\|\mathrm{d}\mu\to0$$ Denote the Bochner integrable functions by $\mathcal{L}_\mathfrak{B}$.

Define the generalized Riemann integral by: $$\int_\mathfrak{R}F\mathrm{d}\mu:=\lim_\mathcal{P}\{\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)\}_\mathcal{P}$$ over finite measurable partitions: $$\mathcal{P}\subseteq\Sigma:\quad\Omega=\bigsqcup_{A\in\mathcal{P}}A\quad(\#\mathcal{P}<\infty)$$ being ordered by refinement: $$\mathcal{P}\leq\mathcal{P}':\iff\forall A'\in\mathcal{P}'\exists A\in\mathcal{P}:\quad A\supseteq A'$$ Denote Riemann integrable functions by $\mathcal{L}_\mathfrak{R}$.

Problem

Clearly, boundedness is a necessary condition for Riemann integrability: $$f:(0,1]\to\mathbb{R}:x\mapsto\frac{1}{\sqrt{x}}:\quad f\in\mathcal{L}_\mathfrak{B},\,f\notin\mathcal{L}_\mathfrak{R}$$

However, there are also bounded Riemann integrable functions that are not Bochner integrable: $$F:[0,1]\to\ell^2[0,1]:t\mapsto\chi_t:\quad F\notin\mathcal{L}_\mathfrak{B},\,F\in\mathcal{L}_\mathfrak{R}$$

Moreover, both strictly include the uniform integral: $$\mathcal{L}_\mathfrak{U}\subsetneq\mathcal{L}_\mathfrak{B}\land\mathcal{L}_\mathfrak{U}\subsetneq\mathcal{L}_\mathfrak{R}$$

So the remaining question is wether at least for the bounded case: $$\|F\|_\infty<\infty:\quad F\in\mathcal{L}_\mathfrak{B}\implies F\in\mathcal{L}_\mathfrak{R}$$ and especially: $$\int_\mathfrak{R}F\mathrm{d}\mu=\int_\mathfrak{B}F\mathrm{d}\mu$$ (Have a guess! ;))

• I have begun to study some things about integration in banach space but for now I don't know much yet. Can you explain better why the function $t\to \chi_t$ is Riemann integrable but not Bochner ? – Ef_Ci Oct 29 '19 at 13:20
• Sure. The function simply fails to be approximable via simple functions: The elements $\chi_t\in\ell^2[0,1]$ for $t\in[0,1]$ define pairwise orthogonal norm one elements and thus span a subspace of uncountable dimension. In particular, there exists no approximation by a sequence of functions with finite dimensional ranges like simple functions. – C-Star-W-Star Oct 31 '19 at 0:28

Proof

Consider a bounded function $\|F\|_\infty<\infty$.

Now, by measurability there exists a simple approximation: $$S_n\in\mathcal{S}:\quad\|F-S_n\|\to0$$ After cutoff and reset it can be chosen to be bounded and decreasing: $$\|S_n\|\leq2\|F\|:\quad\|F-S_n\|\downarrow0$$ Next, denote sufficiently close points by: $$\Omega_n:=\{\|F-S_n\|<\varepsilon\}$$ These are increasing so by continuity from above: $$\mu(\Omega)<\infty:\quad\mu(\Omega_n^\complement)\stackrel{n\to\infty}{\to}0$$ Finally, choose a partition where simple functions are constant: $$S_n=\sum_{k=1}^KS_k\chi(A_k):\quad\mathcal{P}_n:=\{A_1,\ldots,A_K\}$$

Putting all together one obtains the estimate: $$0\leq\|\sum_{\mathcal{P}}F(a)\mu(A)-\sum_{\mathcal{P}'}F(a')\mu(A')\|\leq\left(\|F-S_N\|_{\Omega_N}\mu(\Omega_N)+\|F-S_N\|_{\Omega_N^\complement}\mu(\Omega_N^\complement)\right)\\+\|S_N-S_N\|_\infty\mu(\Omega)+\left(\|F-S_N\|_{\Omega_N}\mu(\Omega_N)+\|F-S_N\|_{\Omega_N^\complement}\mu(\Omega_N^\complement)\right)\\\leq2\left(\|F-S_N\|_{\Omega_N}\mu(\Omega)+3\|F\|_\infty\mu(\Omega_N^\complement)\right)+0<\varepsilon\quad\left(\mathcal{P},\mathcal{P}'\geq\mathcal{P}_{N(\varepsilon)}\right)$$ (I left out maaany details. I hope it is still clear enough.)

Thus it is also Riemann integrable $F\in\mathcal{L}_\mathfrak{R}$.

Moreover, one has an upper bound: $$\int\|F\|\mathrm{d}\mu\leq\|F\|_\infty\mu(\Omega)<\infty$$

Hence it is also Bochner integrable $F\in\mathcal{L}_\mathfrak{R}$.

Besides, the preceding construction gives: $$\|\int_\mathfrak{R}F\mathrm{d}\mu-\int S_n\mathrm{d}\mu\|\leq\ldots\leq\|\int_\mathfrak{R}F\mathrm{d}\mu-\sum_{\mathcal{P}(\varepsilon)}F(a)\mu(A)\|\\+\|F-S_n\|_{\Omega_N}\mu(\Omega)+3\|F\|_\infty\mu(\Omega_N^\complement)<\varepsilon\quad(n\geq N(\varepsilon))$$ and therefore the integrals coincide: $$\int_\mathfrak{R}F\mathrm{d}\mu=\lim_n\int S_n\mathrm{d}\mu=\int_\mathfrak{B}F\mathrm{d}\mu$$

Remark

Note that the above shows for finite measures: $$\|F\|_\infty<\infty:\quad \|F-S_n\|\downarrow0\implies\int S_n\mathrm{d}\mu\to\int_\mathfrak{R}F\mathrm{d}\mu\quad(F\in\mathcal{L}_\mathfrak{R})$$

Caution that it may be not uniform limit: $$F:(0,1]\to\mathcal{H}:F(\frac{1}{n+1}<t\leq\frac{1}{n}):=e_n$$