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Let $\mathbb K$ be $\mathbb R$ or $\mathbb C$. Let $(X, \mathcal M, \mu)$ be a measure space and let $F$ be a Banach space over $\mathbb K$. I would like to define an integral of a function $f:X \rightarrow F$ that satisfies suitable conditions. My method is motivated by Bourbaki's Integration. They assume that $X$ is a locally compact Hausdorff space and $\mu$ is a Radon measure. So mine is sort of a generalization of their definition of the integral.

If $g: X \rightarrow [0, \infty]$ is a non-negative extended real-valued function, we denote by $\int^* g d\mu$ the upper integral of $g$(for the definition of the upper integral, see Lebesgue's monotone convergence theorem for upper integrals).

We denote by $\mathcal F(X, F)$ the set of functions $X\rightarrow F$. It is a vector space over $\mathbb K$. If $f\in \mathcal F(X, F)$, then we denote $N_1(f) = \int^* |f(x)| d\mu(x)$. We denote by $\mathcal F^1(X, F)$ the set $\{f\in \mathcal F(X, F): N_1(f) \lt \infty\}$. Then $\mathcal F^1(X, F)$ is a vector subspace of $\mathcal F(X, F)$. It is easy to see that $N_1$ is a seminorm on $\mathcal F^1(X, F)$. It can be proved by using Lebesgue monotone convergence theorem for upper integrals(Lebesgue's monotone convergence theorem for upper integrals) that $\mathcal F^1(X, F)$ is complete with the seminorm $N_1$(you just mimic the proof of the corresponding theorem in Bourbaki's Integration).

A function $f: X\rightarrow F$ is called a simple function if $f(X)$ is a finite set and $f^{-1}(a) \in \mathcal M$ for all $a\in f(X)$. $f$ is said to have a finite support if $\mu(f^{-1}(a)) \lt \infty$ for all $a\in f(X) - \{0\}$. Let $\mathcal S$ be the set of simple functions of finite support. It is a subspace of $\mathcal F^1(X, F)$. We denote by $\mathcal L^1(X, F)$ the closure of $\mathcal S$ in $\mathcal F^1(X, F)$ with respect to the seminorm $N_1$.

If $f\in \mathcal S$, we can define $\int f d\mu \in F$ in the obvious way. Since $|\int f d\mu| \le \int |f(x)| d\mu(x) = N_1(f)$, the map $\psi: \mathcal S\rightarrow F$ defined by $\psi(f) = \int f d\mu$ is a continuous linear map. Since $\mathcal S$ is dense in $\mathcal L^1(X, F)$ and $F$ is a Banach space, $\psi$ can be extended to a unique continuous linear map $\phi: \mathcal L^1(X, F)\rightarrow F$. We call an element $f \in \mathcal L^1(X, F)$ integrable and call $\phi(f)$ its integral. We denote $\phi(f)$ by $\int f d\mu$ or $\int f(x) d\mu(x)$.

Now I would like to find some criteria to determine when a function $X \rightarrow F$ is integrable. A function $f:X\rightarrow F$ is said to be $\mathcal B(F)$-measurable if $f^{-1}(U) \in \mathcal M$ for all open subsets $U$ of $F$, where $\mathcal B(F)$ is the $\sigma$-algebra generated by all open subsets of $F$.

A function $f:X\rightarrow F$ is said to be Bochner measurable if it satisfies the following condition.

There exists a sequence of simple functions $f_n,n=1,2,\cdots$ defined on $X$ such that $f(x) = \text{lim}_{n\rightarrow\infty} f_n(x)$ almost everywhere on $X$.

It can be proved that an integrable function is Bochner measurable(you just mimic the Bourbaki's proof of the corresponding fact).

Conversely suppose $f$ is a Bochner measurable function. There exists a null set $N$ and a sequence of simple functions $f_n, n = 1, 2, \cdots$ such that $f(x) = \text{lim}_{n\rightarrow \infty} f_n(x)$ for all $x\in X - N$. Since the pointwise limit of a sequence of $\mathcal B(F)$-measurable functions is $\mathcal B(F)$-measurable(see Limit of measurable functions is measurable?), if we redefine $f$ as $f(x) = 0$ for $x \in N$, then $f$ is $\mathcal B(F)$-measurable. Hence the function $x \rightarrow |f(x)|$ is measurable. If $\int |f(x)| d\mu \lt \infty$, it seems to me that $f$ is integrable in our sense.

How do you prove this if it is correct?

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The following proof is highly influenced by the book Vector Measures by Diestel & Uhl.

First off I claim the following assertion. Let $(X, \mathcal M, \mu)$ be a measure space, $E$ a separable normed vector space over $\mathbb R$ or $\mathbb C$, $f: X \rightarrow E$ a $\mathcal B(E)$-measurable function. Then there is a sequence of countably valued $\mathcal B(E)$-measurable functons $f_n, n= 1,2,\cdots$ such that $f$ is the uniform limit of the sequence $\{f_n\}$, i.e. for any $\epsilon\gt 0$, there is an integer $n_0\ge 1$ such that $|f(x) - f_n(x)| \lt \epsilon$ for all $n\ge n_0$ and $x\in X$. The proof goes as follows.

Since $E$ is separable, there exists a countable subset $A = \{a_1, a_2, \cdots\} \subset E$ such that $A$ is dense in $E$. Let $p \gt 1$ be an integer. Define $A_{np} = \{x\in X: |f(x) - a_n| \lt 1/p\}$ for each $n$. Since $f$ is $\mathcal B(E)$-measurable, $A_{np} \in \mathcal M$. Since $A$ is dense in $E$, $X = \bigcup_{n=1}^{\infty} A_{np}$. Define $B_{np} \in \mathcal M$ as follows. Let $B_{1p} = A_{1p}$. If $n \gt 1$ then let $B_{np} = A_{np} - (A_{1p}\cup \cdots \cup A_{{n-1}p})$. Define $f_p = \sum_{n=1}^{\infty} a_n\chi_{B_{np}}$, where $\chi_{B_{np}}$ is the characteristic function of $B_{np}$. We claim that the sequence $\{f_p, p = 1,2,\cdots\}$ is what we are looking for. Clearly $f_p$ is a countably valued $\mathcal B(E)$-measurable function. Choose $\epsilon \gt 0$. There is an integer $n_0 \ge 1$ such that $1/n_0\lt \epsilon$. Then $|f(x) - f_p(x)| \lt 1/p \lt \epsilon$ for all $p\ge n_0$ and $x\in X$ as claimed.

Secondly I claim the following assertion. Let $(X, \mathcal M, \mu)$ be a measure space, $F$ a Banach space over $\mathbb R$ or $\mathbb C$. Let $f: X \rightarrow F$ be a Bochner measurable function. Then there is a sequence of countably valued $\mathcal B(F)$-measurable functons $f_n, n= 1,2,\cdots$ such that $f$ is the almost uniform limit of the sequence $\{f_n\}$, i.e. there is a $\mu$-null set $N$ and for any $\epsilon\gt 0$, there is an integer $n_0\ge 1$ such that $|f(x) - f_n(x)| \lt \epsilon$ for all $n\ge n_0$ and $x\in X - N$.

The proof goes as follows. Since $f$ is Bochner measurable, there is a $\mu$-null set $N$ and a sequence of simple functions $g_n, n = 1,2,\cdots$ such that $f(x) = \text{lim}_{n=1}^{\infty} g_n(x)$ for all $x\in X - N$. Hence $f|(X - N)$ is $\mathcal B(F)$-measurable(as the OP wrote, see Limit of measurable functions is measurable?). Furthermore the closure of $f(X - N)$ in $F$ is a separable subspace of $F$. Hence by what we have just proved, there is a sequence $f_n. n= 1,2, \cdots$ of countably valued $\mathcal B(F)$-measurable functions on $X - N$ such that $f$ is the unform limit of the sequence $\{f_n\}$ on $X - N$. Define a countably valued $\mathcal B(F)$-measurable function $\bar{f_n}$ for each $n$ as follows. Let $\bar{f_n}(x) = f_n(x)$ if $x \in X - N$ and $\bar{f_n}(x) = 0$ if $x \in N$. Then the sequence $\{\bar{f_n})\}$ satisfies our claim.

Now we prove your assertion. Since $\int |f(x)| d\mu \lt \infty$, The set $\{x\in X: f(x) \neq 0\}$ is $\sigma$-finite. Hence we may assume $X$ is $\sigma$-finite. Suppose $X = \bigcup_{n=1}^{\infty} X_n$ is a pairwise disjoint union, where for each $n$ $X_n$ is measurable and $\mu(X_n) \lt \infty$. Suppose for each $n$, $f$ is integrable on $X_n$. Then I claim that $f$ is integrable on $X$. For each $n$, there is a sequence $\{f_{nm}, m = 1,2, \cdots\}$ of simple functions of finite support on $X_n$ such that $\text{lim}_{m\rightarrow \infty} \int_{X_n} |f(x) - f_{nm}(x)| d\mu(x) = 0$. Choose $\epsilon \gt 0$. For each integer $n \ge 1$, there is an integer $p_n$ such that $\int_{X_n} |f(x) - f_{np_n}(x)| d\mu(x) \lt \epsilon/2^n$. We consider $f_{np_n}$ as a step function defined on $X$ by defining $f_{np_n}(x) = 0$ if $x \in X - X_n$. There is an integer $n\ge 1$ such that $\int_{Z_n} |f(x)| d\mu(x) \lt \epsilon$, where $Z_n = \bigcup_{k={n+1}}^{\infty} X_k$. Let $Y_n = \bigcup_{k=1}^n X_k$. Define $g_n = \sum_{k=1}^n f_{kp_k}$. Then $\int_{X} |f(x) - g_n(x)| d\mu(x) = \int_{Y_n} |f(x) - g_n(x)| d\mu(x) + \int_{Z_n} |f(x)| d\mu(x)$ $=\sum_{k=1}^n \int_{X_k} |f(x) - f_{kp_k}(x)| d\mu(x) + \int_{Z_n} |f(x)| d\mu(x) \lt \epsilon + \epsilon$. Hence $f$ is integrable on $X$ as claimed.

Thus we may assume that $\mu(X) \lt \infty$. By the second claim of mine, there are a null set $N$ and a sequence of countably valued $\mathcal B(F)$-measurable functons $f_n, n= 1,2,\cdots$ such that for any $\epsilon\gt 0$, there is an integer $n_0\ge 1$ such that $|f(x) - f_n(x)| \lt \epsilon$ for all $n\ge n_0$ and $x\in X - N$. Since a null set does not affect $\int |f(x)| d\mu(x)$, we may assume $N$ is an empty set. For each integer $n\ge 1$, there is an integer $p_n\ge 1$ such that $|f(x) - f_{p_n}(x)| \lt 1/n$ for all $x\in X$. Let $g_n = f_{p_n}$ for all $n$. Suppose for each integer $n \ge 1$, $g_n = \sum_{k=1}^{\infty} a_{nk}\chi_{A_{nk}}$, where $A_{nk}, k = 1, 2, \cdots$ are pairwise disjoint measurable sets. Then $|g_n(x)| \le |f(x)| + 1/n$ for all $x\in X$. Since $\mu(X) \lt \infty$, $\int |g_n(x)| d\mu(x) \lt \infty$. Since $\int |g_n(x)| d\mu(x) = \sum_{k=1}^{\infty} |a_{nk}|\mu(A_{nk})$, for each $n$, there is an integer $p_n\ge 1$ such that $\int_{B_n} |g_n(x)| d\mu(x) \lt \mu(X)/n$, where $B_n = \bigcup_{k={p_n+1}}^{\infty} A_{nk}$. Set $h_n = \sum_{k=1}^{p_n} a_{nk}\chi_{A_{nk}}$ Then each $h_n$ is a simple function of finite support and $\int |f(x) - h_n(x)| d\mu(x) \le \int_X |f(x) - g_n(x)| d\mu(x) + \int |g_n(x) - h_n(x)| d\mu(x) \le \mu(X)/n + \mu(X)/n$. Therefore $f$ is integrable and the proof is completed.

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