If $L^1$ is the space of lebesgue-integrable functions on $(X, \mathcal{F}, \mu)$, does it matter whether or not $\mu$ is complete? Let $L_1(\mu)$ be the space of Lebesgue-integrable functions on $(X, \mathcal{F}, \mu)$. That is, $L^1$ is the set of equivalence classes of the relation $f \sim g \iff f = g\ \mu-$a.e., equipped with the metric $d(f, g) = \int |f - g|\ d\mu$.
Is it important whether or not $(X, \mathcal{F}, \mu)$ is a complete measure space? What are some properties of $L^1$ that hold only if $\mu$ is complete?
 A: Yes it sometimes depends.
For example, the Dominated Convergence Theorem is saying that if $f_n$ is dominated by $g$ and converges pointwise to $f$ then $f$ is integrable and
$$
\lim_n \int |f_n-f| d \mu =0
$$
When $\mu$ is complete, it is enough to assume that $f_n$ is dominated by $g$ and converges pointwise $\mu$-almost everywhere to $f$, but this is not true in general.
For a counterexample, pick a subset $E \subset X$ which is a subset of a null set but no measurable and observe that $f_n=0$ converges to $f=1_{E}$ $\mu$-almost everywhere.
[Side note I actually run into this issue once in a paper, luckly I could assume that my measure ws complete.]
A: (1) If you have a single measurable function (or countably many measurable functions) then anything you define from them is again measurable.  So completeness is not needed.
(2) But if you have uncountably many, that no longer works.  For example, in probability theory, a "stochastic process" may be a collection $(f_t)_{t \ge 0}$ of random variables indexed by the nonnegative reals.  And you may want some uncountable constructions still to yield something measurable.  That is where completeness is important.  Example:
$$
T(x) = \inf\{t \; : \; |f_t(x)| \ge 1\}
$$
We may want $T$ to be measurable.  In some cases, we can provide natural conditions on the process $(f_t)$ that will insure this when the measure is complete.
(3)  If $E \subseteq \mathbb R^2$ is a Borel set, then the projection onto the $x$-axis is
$$
p(E) = \{x \in \mathbb R\;:\;\exists y, (x,y \in E\} .
$$
If $E$ is Borel in $\mathbb R^2$, then $p(E)$ need not be a Borel set in $\mathbb R$.  But in fact $p(E)$ is measurable with respect to any complete Borel measure on $\mathbb R$.
