How are integrals of non-measurable functions defined in Folland? (Dominated Convergence Theorem)

Question

This is the statement of the LDCT stated in Folland:

The Dominated Convergence Theorem – Let $\{f_n\}$ be a sequence in $L^1$ such that (a) $f_n\rightarrow f$ a.e., and (b) there exists a nonnegative $g\in L^1$ such that $|f_n|\leq g$ a.e. for all $n$. Then $f\in L^1$ and $\int f = \lim_{n\rightarrow \infty}\int f_n$.

It seems like here in order to integrate $f$, we must assume the ambience space to be complete. Otherwise the function $f$ is not measurable. I know that Folland said in the proof that $f$ is measurable up to a null set. However, I am not so sure how this is going to justify the notation "$\int f$". What measure is this integral using?

Results that I think is useful to justify, but not sure how:

Update after reading the comments:

Can I make sense of it this way: We are defining in general the integral $\int_X f := \int_X f \,d\overline{\mu}$ where $f$ is $\overline{\mathcal{M}}$-measurable. Now Prop 2.12 gives us the existence of a $\mathcal{M}$-measurable representative for $f$, we call that representative $g$. Then it make sense to say the following: $\int_X f := \int_X f \,d\overline{\mu} = \int_X g \,d\overline{\mu} = \int_X g \,d\mu$. This chain of equality is always true due to Prop 2.12. Therefore, we can say that when $f$ is only measurable in the completed space and we are integrating it, then we are really integrating the $\mu$-measurable representative using the primal (incomplete) measure $\mu$.

Answer 1

In general, I'd prefer to avoid going to the completion if possible. It is possible to reword the theorem slightly as follows:

Dominated Convergence Theorem.

Let $\{f_n\}_{n=1}^{\infty}$ be a sequence in $L^1(X,\mathfrak{M},\mu)$ such that

• (a) $\{f_n\}_{n=1}^{\infty}$ converges pointwise a.e on $X$.
• (b) there is a $g\in L^1$ such that for all $n\in\Bbb{N}$, $|f_n|\leq g$ a.e.

Then, for every measurable function $f$ for which $f_n\to f$ a.e, we have that $f\in L^1(\mu)$ and $\int_X f\,d\mu=\lim\int_X f_n\,d\mu$.

Note that the conclusion is not vacuous since there does indeed exist such a measurable $f$ (and this is what Folland alludes to, though in a roundabout manner), because of the following simple lemma:

Lemma. Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of measurable (complex-valued) functions on $(X,\mathfrak{M},\mu)$ which converges pointwise a.e to some function $\phi$. Then, there is a measurable function $f$ on $X$ such that $f_n\to f$ a.e.

To prove this, note that by hypothesis, there is an $N\in\mathfrak{M}$ such that $\mu(N)=0$ and for all $x\in X\setminus N$, we have $f_n(x)\to \phi(x)$. Define $f:X\to\Bbb{C}$ as $f(x):=\lim\limits_{n\to\infty}(f_n\chi_{N^c})(x)$. Note that each $f_n\chi_{N^c}$ is a measurable function, and this sequence of functions converges pointwise everywhere (if $x\in N^c$ then the sequence converges to $\phi(x)$ by hypothesis, and otherwise to $0$). So, $f$ being the pointwise limit of measurable functions is again measurable, and clearly $f_n\to f$ on $N^c$, and hence pointwise a.e. This completes the proof.

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What you suggested (introducing extra notation for integrating non-measurable functions) seems fine, but I'm hesitant to pass to the completion unless I really have to, and I'd rather not make new definitions for existing symbols unless it's necessary. In this case, it's not necessary to go this route, since as shown above, it's no loss of generality in assuming $f$ is measurable from the outset. Also, Folland's exact words in the proof were

$f$ is measurable (perhaps after redefinition on a null set)

so clearly he intended such a redefinition as in the lemma (he mentions proposition 2.12, but I honestly do not see the need for that here, as argued above).