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Call $f:I\to F$ gauge integrable where $I = [a, b]$ is a compact interval and $F$ is a Banach space, if the usual definition holds like if $F = \mathbb{R}$, just replace absolute value by norm. How can one prove the following?

Theorem 1. If $f:I\to F$ is gauge integrable, $I = [a, b]$ then its indefinite integral $F(x) := \int_a^x f$ is differentiable a.e.

Above theorem might not be true, it should be true when $F$ is finite-dimensional.

Lemma 1 (Saks-Henstock). Let $f:I\to F$ be gauge integrable, $\varepsilon > 0$ and $\delta$ be a gauge such that for all $\delta$-fine paritions $\mathcal{P}$ $$\left\|\int_I f - S(f, \mathcal{P})\right\|\leq \varepsilon.$$ If $\mathcal{Q} = \{(I_i, t_i)\}$ is a $\delta$-fine subpartition of $I$, then $$\left\| \sum_i \int_{I_i} f - S(f, \mathcal{Q})\right\| \leq \varepsilon.$$

Above lemma is fine as it is for any Banach space $F$. Here $S(f, \mathcal{P})$ is the Riemann sum of $f$ with respect to (sub)partion $\mathcal{P}$.

Lemma 2. If $f, \varepsilon$ and $\delta$ are like above, $F = \mathbb{R}$ then $$\sum_i \left\|\int_{I_i} f - f(t_i)\ell(I_i)\right\| \leq 2\varepsilon.$$

Here $\ell(J)$ is length of $J$, and for $F = \mathbb{C}$ lemma 2 holds with $2$ replaced by $4$.

The proof of theorem 1 for $F = \mathbb{R}$ uses Vitali's covering lemma and lemma 2, but as this seems to be unavailable when $F$ is an infinite-dimensional Banach space, how can I prove theorem 1 in this case, and is it even true?

Jakobian
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1 Answers1

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The answer is no, even for Riemann integrable functions. Let $F$ be the space of bounded functions from $[0, 1]$ to $\mathbb{R}$ with supremum norm, and let $T:[0, 1]\to F$ be defined as $T(r) = 1_{[r, 1]}$, then $T$ is Riemann integrable, with integral $y(t) = t$, $y\in F$. In particular, its gauge integrable. But its indefinite integral $G(x) = \int_0^x T$, where $G(x)(t) = \min(x, t)$ is not differentiable at any point.

Note that $T$ is not Bochner integrable, thus its also an example of Riemann integrable function thats not Bochner integrable.

Jakobian
  • 15,280