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?