Recently, I've come across this interesting equivalence of integrability criteria for Bochner integral whose related definitions can be found here. Could you verify if my proof is fine?
Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, | \cdot |)$ a Banach space. For $f \in \mathcal{L}_{0}(X, \mu, E)$, the following are equivalent:
(i) $f \in \mathcal{L}_{1}(X, \mu, E)$;
(ii) $|f| \in \mathcal{L}_{1}(X, \mu, \mathbb{R})$;
(iii) $\int_{X} |f| \mathrm d \mu<\infty$ in the sense of Lebesgue integral.
Proof:
- (i) $\to$ (ii)
Let $(f_n)$ is a Cauchy sequence in $\mathcal S (X, \mu, E)$ such that $f_n \to f$ $\mu$-a.e. We have $\big | |f_n| - |f_m| \big| \le |f_n - f_m|$, so $$\big \| |f_n| - |f_m| \big \|_1 = \int_X \big | |f_n| - |f_m| \big| \mathrm d \mu \le \int_X |f_n - f_m|\mathrm d \mu = \|f_n - f_m \|_1.$$
Hence $(|f_n|)$ is a Cauchy sequence in $\mathcal S (X, \mu, \mathbb R)$ such that $|f_n| \to |f|$ $\mu$-a.e. Hence $|f| \in \mathcal{L}_{1}(X, \mu, \mathbb{R})$.
- (ii) $\to$ (iii)
Let $(f_n)$ is a Cauchy sequence in $\mathcal S (X, \mu, \mathbb R)$ such that $f_n \to |f|$ $\mu$-a.e. By definition, $\int_X |f| \mathrm d \mu := \lim_n \int_X f_n \mathrm d \mu$. We have $$\left | \int_X f_n \mathrm d \mu - \int_X f_m \mathrm d \mu \right | \le \int_X | f_n - f_m| \mathrm d \mu = \|f_n-f_m \|_1.$$
So $(\int_X f_n \mathrm d \mu)_n$ is a Cauchy sequence in $\mathbb R$ and thus converges in $\mathbb R$. Hence $\int_X |f| \mathrm d \mu < \infty$.
- (iii) $\to$ (ii)
Suppose $(f_n)$ is a non-decreasing sequence in $\mathcal S (X, \mu, \mathbb R^+)$ such that $f_n \to |f|$ $\mu$-a.e. Then $\int_X f_n \mathrm d \mu \nearrow \int_X |f| \mathrm d \mu$. Then $\int_X |f| \mathrm d \mu < \infty$ implies for $\varepsilon >0$, there is $N \in \mathbb N$ such that $\big | \int_X f_n \mathrm d \mu -\int_X |f| \mathrm d \mu \big | = \int_X \big | |f| - f_n \big |\mathrm d \mu < \varepsilon/2$ for $n \ge N$. It follows that $$\| f_n - f_m \|_1 = \int_X |f_n-f_m| \mathrm d \mu \le \int_X \big | |f| - f_n \big | \mathrm d \mu + \int_X \big | |f| - f_m \big | \mathrm d \mu \le \varepsilon.$$
Hence $(f_n)$ is a Cauchy sequence in $\mathcal S (X, \mu, \mathbb R^+)$ and thus $|f| \in \mathcal{L}_{1}(X, \mu, \mathbb{R})$.
- (ii) $\to$ (i)
Let $\left(f_{n}\right)$ be a sequence in $\mathcal{S}(X, \mu, E)$ such that $f_{n} \to f$ $\mu$-a.e. Let $A_{n}:= \{x \in X \mid \left|f_{n}(x)\right| \le 2 |f(x)| \}$ and $g_{n} := 1_{A_{n}} f_{n}$ for $n \in \mathbb N$. Then $\left(g_{n}\right)$ is a sequence in $\mathcal{S}(X, \mu, E)$. Let's prove that $(g_n)$ converges $\mu$-a.e. to $f$.
Let $A := \{x \in X | f (x) \neq 0 \}$ and $B$ be a null set such that $f_n (x) \to f (x)$ for all $x \in A^c$.
For each $A \cap B^c$, there is $N \in \mathbb N$ such that $|f_n(x) - f (x)| \le |f(x)|$, and thus $|f_n(x)| \le 2 |f(x)|$, and thus $x \in A_n$, and thus $1_{A_n} (x) = 1$, and thus $g_n (x) = f_n (x)$ for $n \ge N$. This means $g_n (x) \to f(x)$ for $x \in A \cap B^c$.
For $x \in A^c \cap B^c$, $f(x) = 0$ and thus $|g_n (x) - f(x)| = |g_n(x)| = 1_{A_{n}} |f_{n}(x)| \le |f_n(x)|$. It follows from $f_n (x) \to 0$ that $g_n (x) \to 0$. Hence $g_n \to f$ on $A^c \cap B^c$.
Clearly, $|f - g_n| \to 0$ $\mu$-a.e. and $|f - g_n| \le 3 |f|$ for all $n \in \mathbb N$. By "reverse" Fatou’s lemma, we have $$\limsup_n \int_X |f - g_n| \mathrm d \mu \le \int_X \limsup_n |f - g_n| \mathrm d \mu = 0.$$
Hence $\int_X |f - g_n| \mathrm d \mu \to 0$ as $n \to \infty$. Thus means for $\varepsilon>0$, there is $N \in \mathbb N$ such that $\int_X |f - g_n| \mathrm d \mu<\varepsilon/2$ for all $n \ge N$. On the other hand, $$\|g_n - g_m \|_1 = \int_X |g_n - g_m| \mathrm d \mu \le \int_X |f - g_n| \mathrm d \mu + \int_X |f - g_m| \mathrm d \mu < \varepsilon.$$
Hence $(g_n)$ is a Cauchy sequence in $\mathcal{S}(X, \mu, E)$ and thus $f \in \mathcal{L}_1(X, \mu, E)$.
