Let $(X,\mathcal{A},\mu)$ be a measure space. Then define a set $N \subseteq X$ (globally) null if $\mu(N) = 0$ and locally null if $\forall A \in \mathcal{A}$ with $\mu(A) < \infty$: $\mu(A \cap N) = 0$.
Firstly, note that a globally null set is also locally null and if $\mu$ is sigma finite, the two notions coincide. Thus a function that vanishes globally a.e. also vanishes locally a.e., and again, if the measure is sigma finite, the two coincide.
proof: Let $N$ be globally null and let $A \in \mathcal{A}$ with $\mu(A) < \infty$ be arbitrary, then $\mu(A \cap N) \leq \mu(N) = 0$.
proof: Let $\mu$ be sigma finite and let $N$ be locally finite. Let $A_n \uparrow X$ be an nested sequence of sets $A_n \in \mathcal{A}$ s.t. $\forall n \in \mathbb{N}: \mu(A_n) < \infty$. Then
$$ \mu(N) = \mu(\cup_{n = 1}^{\infty} A_n \cap N) \leq \sum_{n = 1}^{\infty} \mu(A_n \cap N) = 0$$
Furthermore, a measurable locally null set $N$ that has finite measure is also globally null since
$$0 = \mu(N \cap N) = \mu(N)$$
I am looking for an of a $p$-integrable function which vanishes locally a.e. but not globally a.e.
For functions in $L^{\infty}(X)$ there are immediate examples. Let $N$ be a set which is locally $0$ but not globally, then $1_N$ vanishes locally a.e. but not globally a.e.
However I don't know about $p$ integrable functions for $p < \infty$ and I suspect that any $p$-integrable function which is locally $0$ is also globally $0$.
The motivation for this question comes from the dual isometries of $L^p$ spaces:
Let $1 < p < \infty$ and $q > 0$ s.t. $\frac{1}{p} + \frac{1}{q} = 1$. Then $L^q(X) \rightarrow (L^p(X))^*$ via $g \mapsto \int – \cdot g d\mu$ is an injective isometry. And when $\mu$ is sigma finite, this is also true for $p = 1$ (and $q = \infty$).
For this theorem $L^p$ is defined by identifying functions that differ on globally null sets and for $q < \infty$) this works fine, but in order to salvage the theorem for the non sigma finite case, one needs to redefine $L^{\infty}(X)$ to identify functions that differ on locally null sets.
Since for the sigma finite case the two notions coincide, we may generally define $L^{\infty}(X)$ by identifying functions differing on locally null sets. But then how does that work for $p < \infty$? If we also want to generally define $L^p(X)$ by locally null, then since the previously mentioned embedding of $L^q(X)$ into $(L^p(X))^*$ is injective, functions differing only locally on null sets should also differ only on globally null sets.
Best Answer
If $\int|f|^p < \infty$, then $\{x : f(x) \ne 0\} = \bigcup_n A_n$ where $A_n = \{x:|f(x)| > \frac1n\}$. But $A_n$ has finite measure.