Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $E := [0, \infty]$ endowed with order topology.
$f \in E^{X}$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $e_k \in E \setminus \{\infty\}$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. The Bochner integral of such $f$ w.r.t. $\mu$ is defined by $\int f := \sum_{k=1}^n e_k \mu(A_k)$. Let $\mathcal S (X, \mu, E)$ be the space of $\mu$-simple functions.
$f \in E^{X}$ is called $\mathcal A$-measurable if $f^{-1} (O) \in \mathcal A$ for every open subset $O$ of $E$.
$f \in E^{X}$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n) \subset \mathcal S (X, \mu, E)$. Let $\mathcal L_0 (X, \mu, E)$ be the space of such $\mu$-measurable functions.
Lemma 1: $f \in E^X$ is $\mu$-measurable if and only if there is an increasing sequence $(f_n) \subset \mathcal S (X, \mu, E)$ such that $f_n \nearrow f$ $\mu$-a.e.
For a $\mu$-measurable function $f \in E^X$, by Lemma 1 there is an increasing sequence $(f_n) \subset \mathcal S (X, \mu, E)$ such that $f_n \nearrow f$ $\mu$-a.e. Then Lebesgue integral of such $f$ is defined as $\oint f := \lim_n \int f_n$ and is independent of the approximating sequence $(f_n)$.
$f \in E^{X}$ is called $\mu$-integrable if $f$ is a $\mu$-a.e. limit of a Cauchy sequence $(f_n) \subset \mathcal S (X, \mu, E)$. The Bochner integral of such $f$ w.r.t. $\mu$ is defined by $\int f := \lim_n \int f_n$ and is independent of the approximating sequence $(f_n)$.
Lemma 2: Let $f,g \in E^X$ be $\mu$-integrable such that $f \le g$ $\mu$-a.e. Then $\int f \le \int g$.
- Let $\mathcal L_1 (X, \mu, E)$ be the space of $\mu$-integrable functions. Clearly, $\mathcal L_1 (X, \mu, E) \subset \mathcal L_0 (X, \mu, E)$.
I'm trying to prove the equivalence between these two types of integrals
Theorem: If $f \in \mathcal L_1 (X, \mu, E)$ then $\int f = \oint f$. If $f \in \mathcal L_0 (X, \mu, E)$ such that $\oint f < \infty$ then $f \in \mathcal L_1 (X, \mu, E)$ and $\int f = \oint f$.
Could you have a check on my proof?
My attempt:
- Let $f$ be $\mu$-integrable.
Then $f$ is $\mu$-measurable. There is a sequence $(f_n) \subset \mathcal S (X, \mu, E)$ such that $f_n \nearrow f$ $\mu$-a.e. We have $\oint f = \lim_n \int f_n$. It suffices to show that $(f_n)$ is a Cauchy sequence. It follows from $f_n \le f$ and Lemma 2 that $\int f_n \le \int f$. Taking the limit $n \to \infty$, we get $\oint f_n \le \int f <\infty$. Fix $\varepsilon>0$, there is $N$ such that $$ \oint f - \int f_n=\left | \oint f - \int f_n\right | <\varepsilon \quad \forall n >N. $$
Then $$ \|f_n-f_m\|_1 = \int |f_n-f_m| = \int f_m - \int f_n \le \oint f - \int f_n < \varepsilon \quad \forall m \ge n > N. $$
It follows that $(f_n)$ is a Cauchy sequence.
- Let $\oint f < \infty$.
Let $(f_n) \subset \mathcal S (X, \mu, E)$ such that $f_n \nearrow f$ $\mu$-a.e. Then $\oint f = \lim_n \int f_n < \infty$. Fix $\varepsilon>0$, there is $N$ such that $$ \oint f - \int f_n=\left | \oint f - \int f_n\right | <\varepsilon \quad \forall n >N. $$
Then $$ \|f_n-f_m\|_1 = \int |f_n-f_m| = \int f_m - \int f_n \le \oint f - \int f_n < \varepsilon \quad \forall m \ge n > N. $$
It follows that $(f_n)$ is a Cauchy sequence. Then $\int f =\lim_n \int f_n$. This completes the proof.
