I'm trying to prove dominated convergence theorem for Banach space. Could you verify if my proof is fine or contains some subtle mistakes?
Let $(f_n)$ be a sequence in $\mathcal L_0 (X, \mu, E)$. Suppose that there exists $g \in \mathcal L_1 (X, \mu, E)$ such that $|f_n| \le |g|$ $\mu$-a.e. for all $n$. Suppose also that, for some $f \in E^X$, $f_n \to f$ $\mu$-a.e. Then $f_n,f \in \mathcal L_1 (X, \mu, E)$ for all $n$, $f_n \to f$ in $\mathcal L_1 (X, \mu, E)$, and $\int_X f_n \mathrm d \mu \to \int_X f \mathrm d \mu$ in $E$.
First, we recall related definitions.
Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, | \cdot |)$ a Banach space.
A function $f \in E^{X}$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $0 \neq e_k \in E$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. The integral of such $f$ w.r.t. $\mu$ is defined by $\int_{X} f \mathrm d \mu := \sum_{k=1}^n e_k \mu(A_k)$.
Let $\mathcal S (X, \mu, E)$ be the space of such $\mu$-simple functions. We define $\| \cdot \|_1 : \mathcal S (X, \mu, E) \to \mathbb R$ by $f \mapsto \int_{X} |f| \mathrm d \mu$. Then $\| \cdot \|_1$ is a semi-norm on $\mathcal S (X, \mu, E)$. The notion of Cauchy sequence is thus applicable for $(\mathcal S (X, \mu, E), \| \cdot \|_1)$.
A function $f \in E^{X}$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n)$ in $\mathcal S (X, \mu, E)$. Let $\mathcal L_0 (X, \mu, E)$ be the space of such $\mu$-measurable functions.
A function $f \in E^{X}$ is called $\mu$-integrable if $f$ is $\mu$-a.e. limit of a Cauchy sequence $(f_n)$ in $\mathcal S (X, \mu, E)$. The integral of such $f$ w.r.t. $\mu$ is defined by $\int_{X} f \mathrm d \mu := \lim_n \int_{X} f_n \mathrm d \mu$. We also define the semi-norm $\| f\|_1 := \int_{X} |f| \mathrm d \mu$.
Let $\mathcal L_1 (X, \mu, E)$ be the space of such $\mu$-integrable functions. Then $\big (\mathcal L_1 (X, \mu, E), \| \cdot \|_1 \big )$ is complete. Also, $\mathcal S (X, \mu, E)$ is dense in $\mathcal L_1 (X, \mu, E)$ w.r.t. $\| \cdot \|_1$.
Proof: First, we assume $f_n \in \mathcal L_1 (X, \mu, E)$ for all $n$. Let $0 \le g_n := \sup_{i,j \ge n} |f_i - f_j| \le 2 |g|$. Then $(g_n)$ is a non-increasing sequence in $\mathcal L_0 (X, \mu, \mathbb R^+)$ that converges to $0$ $\mu$-a.e. By "reverse" Fatou's lemma, $$0 \le \limsup_n \int_X g_n \mathrm d \mu \le \int_X \limsup_n g_n \mathrm d \mu = 0.$$
It follows that $\int_X g_n \mathrm d \mu \to 0$ in $\mathbb R^+$. For all $i, j \ge n$, $$\|f_i -f_j \|_1 = \int_X |f_i - f_j| \mathrm d \mu \le \int_X g_{n} \mathrm d \mu.$$
Hence $(f_n)$ is a Cauchy sequence in $\mathcal L_1 (X, \mu, E)$ which is complete. Then $f_n$ converges to some $\hat f$ in $\mathcal L_1 (X, \mu, E)$.
Second, we will show that there is a subsequence $(f_{\varphi(\ell)})$ of $(f_n)$ such that $f_{\varphi(\ell)} \to \hat f$ $\mu$-a.e as $\ell \to \infty$.
It suffices to assume $\hat f =0$, because, if $\hat f \neq 0$, we can consider the sequence $(f_n - \hat f)$.
There is a subsequence $\varphi$ of $(n)$ such that $\|f_i-f_j\|_1 \le 2^{-2\ell}$ for $i,j \ge \varphi(\ell)$. Let $h_\ell :=f_{\varphi(\ell)}$. Then $\|h_\ell - h_m\|_1 \le 2^{-2\ell}$ for $m \ge \ell$. Take the limit $m \to \infty$, we get $\|h_\ell\|_1 \le 2^{-2\ell}$.
Let $B_\ell := \{ x\in X \mid |h_\ell (x)| \ge 2^{-\ell} \}, A_n := \cup_{\ell \ge n} B_\ell$, and $A := \cap A_n$. Then $\mu(B_\ell) \le 2^{-\ell}$ for $\ell \in \mathbb N$, $\mu(A_n) \le 2^{-n+1}$ for $n \in \mathbb N$, and $\mu(A) =0$. Then $(h_\ell)$ converges to $0$ uniformly on each $A_n^c$ and pointwise on $A^c$.
On the other hand, $f_n \to f$ $\mu$-a.e., so does $(f_{\varphi(\ell)})$. Hence $f = \hat f$ $\mu$-a.e. and thus $f_n$ converges to $f$ in $\mathcal L_1 (X, \mu, E)$. Next we have $$\left |\int_X f_n \mathrm d \mu - \int_X f \mathrm d \mu \right | = \left | \int_X (f_n - f) \mathrm d \mu \right | \le \int_X |f_n - f| \mathrm d \mu = \|f_n -f\|_1 \to 0.$$
Next we're going to prove $f_n \in \mathcal L_1 (X, \mu, E)$. It suffices to prove for $n=0$. Let $\left(\varphi_{n}\right)$ be a sequence in $\mathcal{S}(X, \mu, E)$ such that $\varphi_{n} \to f_0$ $\mu$-a.e. Let $A_{n}:= \{x \in X \mid \left|\varphi_{n}(x)\right| \le |g(x)| \}$ and $g_{n} := 1_{A_{n}} \varphi_{n}$ for $n \in \mathbb N$. Then $\left(g_{n}\right)$ is a sequence in $\mathcal{S}(X, \mu, E) \subseteq \mathcal{L}_1(X, \mu, E)$ that converges $\mu$-a.e. to $f_0$. Because $\left|g_{n}\right| \le g$ for $n \in \mathbb{N}$, the claim follows from result proved above. This completes the proof.
Update: I have found that the proof of $f_n \in \mathcal L_1 (X, \mu, E)$ is more subtle than I thought. Below is a proper treatment.
Next we're going to prove $f_n \in \mathcal L_1 (X, \mu, E)$. It suffices to prove for $n=0$. Let $\left(\varphi_{n}\right)$ be a sequence in $\mathcal{S}(X, \mu, E)$ such that $\varphi_{n} \to f_0$ $\mu$-a.e. Let $A_{n}:= \{x \in X \mid \left|\varphi_{n}(x)\right| \le 2 |g(x)| \}$ and $g_{n} := 1_{A_{n}} \varphi_{n}$ for $n \in \mathbb N$. Then $\left(g_{n}\right)$ is a sequence in $\mathcal{S}(X, \mu, E) \subseteq \mathcal{L}_1(X, \mu, E)$. Let's prove that $(g_n)$ converges $\mu$-a.e. to $f_0$.
Let $A$ be a null set such that $\varphi_n (x) \to f_0 (x)$ for all $x \in A^c$. Let $B$ be a null set such that $|f_0 (x)| \le |g(x)|$ for $x \in B^c$. Also, $C := A \cup B$ and $D := \{x \in X | g (x) \neq 0 \}$. Then $C$ is also a null set.
For each $x \in C^c \cap D$, there is $N \in \mathbb N$ such that $|\varphi_n(x) - f_0 (x)| \le |g(x)|$ and thus $|\varphi_n(x)| \le |f_0 (x)| + |g(x)| \le 2 |g(x)|$ and thus $x \in A_n$ for $n \ge N$. This means $g_n (x) = \varphi_n (x) \to f_0(x)$ for $x \in C^c \cap D$.
For $x \in C^c \cap D^c$, $g(x) = f_0(x) = 0$ and thus $|g_n (x) - f_0(x)| = |g_n(x)| = 1_{A_{n}} |\varphi_{n}| \le |\varphi_n(x)| \to 0$ and thus $g_n (x) \to f_0(x)$. Hence $g_n \to f_0$ on $C^c$.
Because $\left|g_{n}\right| \le 2|g|$ for $n \in \mathbb{N}$, the claim follows from result proved above. This completes the proof.
