Bochner Lp is complete for a Banach value space

The Cauchy sequence has a subsequence $(f_k)$ with $\mathrm{\parallel }{f}_{k+1}-f_k{\mathrm{\parallel }}_{L^p(X)}\le 2^{-k}$ for $k\ge 1$, by recursively taking the least eligible indices. Countable Choice [A1] selects strongly measurable representatives for these classes. Put $g_N(t)={\sum }_{k=1}^N\mathrm{\parallel }{f}_{k+1}(t)-f_k(t){\mathrm{\parallel }}_X$. By [F3], $g_N$ is measurable; Minkowski [F4] gives $\mathrm{\parallel }{g}_N{\mathrm{\parallel }}_{L^p}\le {\sum }_{k=1}^N{2}^{-k}\le 1$. Monotone convergence [F5] gives $g(t)={\sum }_{k\ge 1}\mathrm{\parallel }{f}_{k+1}(t)-f_k(t){\mathrm{\parallel }}_X$ with $\int g^p\le 1$. Thus $g<\mathrm{\infty }$ almost everywhere: for $A=\{g=\mathrm{\infty }\}$, $M^p{\mathbf{1}}_A\le g^p$ for every $M>0$, and [F6] gives $M^p\mu (A)\le 1$, forcing $\mu (A)=0$.

Outside the null set where $g=\mathrm{\infty }$, the sequence $f_k(t)$ is Cauchy in $X$ and converges to $f(t)\in X$ by [F1]; set $f=0$ on that null set. For each $k$, choose a simple sequence $s_{k,m}\to f_k$ almost everywhere, as supplied by [F2] and [A1]. The sets $E_{k,m}=\{t:\mathrm{\parallel }{s}_{k,m}(t)-f_k(t){\mathrm{\parallel }}_X>2^{-k}\}$ are measurable by [F3], and their indicators tend to zero almost everywhere as $m\to \mathrm{\infty }$. Dominated convergence [F6] on the finite interval [F8] lets Countable Choice [A1] select $m_k$ with $\mu (E_{k,m_k})\le 2^{-k}$. By [F7], the set $N={\bigcap }_{j\ge 1}{\bigcup }_{k\ge j}{E}_{k,m_k}$ has measure at most ${\sum }_{k\ge j}{2}^{-k}\le 2^{1-j}$ for every $j$, hence is null. Outside $N$ and the previous null set, $s_{k,m_k}(t)\to f(t)$, so $f$ is strongly measurable by [F2]. Also $\mathrm{\parallel }f-f_1{\mathrm{\parallel }}_X\le g$ almost everywhere; [F6] makes this difference $L^p$, and Minkowski [F4] gives $f\in L^p$. For each $N$, $\mathrm{\parallel }f(t)-f_N(t){\mathrm{\parallel }}_X\le {\sum }_{k\ge N}\mathrm{\parallel }{f}_{k+1}(t)-f_k(t){\mathrm{\parallel }}_X\to 0$ and is bounded by $g(t)$. Dominated convergence [F6] gives $\mathrm{\parallel }f-f_N{\mathrm{\parallel }}_{L^p(X)}\to 0$.

The original sequence is Cauchy and has the subsequence $f_N\to f$; given $\epsilon >0$, choose $N$ with $\mathrm{\parallel }{f}_N-f{\mathrm{\parallel }}_{L^p(X)}<\epsilon \mathrm{/}2$ and then take the original index sufficiently far out that its distance to $f_N$ is below $\epsilon \mathrm{/}2$. The triangle inequality [F4] proves convergence of the entire sequence to $f$, so $L^p(0,T;X)$ is complete.