Claim: If two functions $f,g: \Omega \to \overline{\mathbb{R}}$ are equal almost everywhere in $E$, i.e. $f=g$ on $E \setminus N$ with $\mu(N)=0$, and $f \in \mathcal{L}(E)$, then $g \in \mathcal{L}(E)$ and the integrals have the same value.
My definitions are as follows:
Definition: Let $(\Omega,\Sigma,\mu)$ be a measure space and $E \in \Sigma$. Then the integral of a non-negative simple function $s: \Omega \to [0,\infty]$ with representation $s(x) = \sum \limits_{j=1}^{k} \alpha_j \chi_{A_j(x)}$ on $E$ is
$$\int_E s d\mu := \sum \limits_{j=1}^{k} \alpha_j \mu(A_j \cap E).$$
Note that it can be shown that the integral does not depend on the representation of $s$.
Definition: Let $(\Omega,\Sigma,\mu)$ be a measure space and $E \in \Sigma$, then the integral of a non-negative function $f: \Omega \to [0,\infty]$ is
$\int_E f d\mu := sup \{\int_E s d\mu: 0 \leq s \leq f \text{ on E, s simple}\}.$
Note that $f$ is not required to be measurable, but integral functions in the end will be required to be measurable.
Definition: Let $(\Omega,\Sigma,\mu)$ be a measure space, $E \in \Sigma$ and $f: \Omega \to \overline{\mathbb{R}}$.
(i) If at least one of the integrals of $\int_E f_+ d\mu$, $\int_E f_- d\mu$ is finite, then the integral of $f$ is
$$\int_E f d\mu := \int_E f_+ d\mu$ - \int_E f_- d\mu.$$
(ii) If $f$ is measurable and both integrals are finite, i.e. $\int_E f d\mu \in \mathbb{R}$, then $f$ is Lebesgue-integrable ($f \in \mathcal{L}(E)$).
Now my idea to prove the claim is as follows:
There is a lemma in my lecture notes that states:
Lemma: Let $(\Omega,\Sigma,\mu)$ be a measure space, $E \in \Sigma$ and $f,g: \Omega \to \overline{\mathbb{R}}$.
$A \subset E, A \in \Sigma$ with $\mu(E \setminus A)=0$ and $f \in \mathcal{L}(E) \implies f \in \mathcal{L}(A)$ and $\int_A f d\mu = \int_E f d\mu.$
The proof proceeds step-wise for simple, non-negative and then general functions.
It is pretty easy to see that the integral equality holds for step functions and then follows for non-negative functions:
A simple function with $0 \leq s \leq f$ on E, also satisfies this condition on $A$, so $\int_A f d\mu \geq \int_E f d\mu$.
A simple function with $0 \leq s \leq f$ on A can be extended to $E$ with $s(x)=0$ $\forall x \in E \setminus A$, so $\int_A f d\mu \leq \int_E f d\mu.$
Since $f \in \mathcal{L}(E)$, both positive and negative part have finite integrals, so $f \in \mathcal{L}(A)$ and the integral equality follows from the one for positive and negative parts.
However, the proof shows that the lemma also holds for non-negative functions without the need to assume that $f \in \mathcal{L}(E)$ and I think we can use this to show the claim if we already know that $g$ is measurable:
$f=g$ on $E \setminus N$ implies $f_{\pm}=g_{\pm}$ on $E \setminus N$.
So $\int_E f_{\pm} = \int_{E \setminus N} f_{\pm} = \int_{E \setminus N} g_{\pm} = \int_E g_{\pm}$, which proves that $g \in \mathcal{L}(E)$ and that the integrals have the same value. For the last step we obviously need $f \in \mathcal{L}(E)$, but the lemma can be used even before we know that $g \in \mathcal{L}(E)$.
Now it remains to show that $g$ is in fact measurable if $f \in \mathcal{L}(E)$. If the measure space is complete, then the measurability of $g$ follows from the measurability of $f$ and we are done (see here).
Now I have two questions:
- Is the completeness of the space necessary for the result to be correct?
- Am I correct that we only need completeness to show that $g$ is measurable, i.e. is my proof correct?
Thanks a lot!
