If $A$ is a Borel set and $\overline A\setminus A$ is a null set, can we embed $L^p(A)$ into $L^p(\overline A)$?

Question

Let $p\ge1$ and $A\in\mathcal B(\mathbb R)$ and assume that $N:=\overline A\setminus A$ is a Leesgue null set.

Can any function $f\in\mathcal L^p(A)$ be extended to a function $g\in\mathcal L^p(\overline A)$?

Since $N\in\mathcal B(\mathbb R)$, $$g(x):=\left.\begin{cases}f(x)&\text{, if }x\in A\\0&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x\in\overline A$$ should be piecewie Borel measurable, hence Borel measurable. And since $N$ is a Lebesgue null set, $\left\|g\right\|_{L^p(\overline A)}=\left\|f\right\|_{L^p(A)}$. Am I missing something?

Remark: The question came to my mind as I've noticed that a function $f:[a,b]\to\mathbb C$ is called absolutely continuous if there is a $h\in L^1((a,b))$ with $f(x)=f(a)+\int_0^xh(y)\:{\rm d}y$ for all $x\in[a,b]$ and it seems akward to me why one is explicitly assuming that $h$ is defined on the open interval $(a,b)$ and not $h\in L^1([a,b])$. Is there anything I'm missing? Even uniqueness (in an "almost everywhere" sense) shouldn't be a problem, since $\{a,b\}$ is a null set.

Answer 1

If the $\mathcal{L}^{p}$ does not mean to be the equivalent class, then the extension is definitely not unique, for you can set $g_{a}(x)=a$ for $x\in N$ and any real number $a$.

And there is no need to look only at the Borel measurable functions. While $\mathcal{L}^{p}$ consists of Lebesgue measurable functions which are $p$-th integrable.

Answer 2

Yes, we can extend any function $f$ by any nullset, and prescribe whatever values we wish in the extension (ok, keep measurablitity), it won't effect its integral, since integral of any function over a nullset is zero, and integrals on disjoint domains add up.

Note moreover, that the elements of $\mathcal L^p(X)$ are already equivalence classes of functions over $X$, so that two functions are equivalent iff they agree almost everywhere.
In lights of this, we will even get effectively the same element of $\mathcal L^p(A\cup M)$ whatever extension on $M$ we take if $M$ has zero measure.