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If $(X,mathcalA,mu)$ is a measure space,
$(X,mathcalB,overlinemu)$ is complete,
$mathcalBsupsetmathcalA$, and $overlinemu(A)=mu(A)$ for
every $AinmathcalA$, is $(X,mathcalB,overlinemu)$
necessarily the completion of $(X,mathcalA,mu)$?

My definition of completion is:

The completion of $mathcalA$ is the smallest $sigma$-algebra $mathcalB$ containing $mathcalA$ such that $(X,mathcalB,mu)$ is complete.

It seems like the answer to my question is no, because it isn't clear that $mathcalB$ is necessarily the smallest $sigma$-algebra satisfying the required properties. But I have been looking at the Completion Theorem where they seem to assume that the answer to my question is yes. How can I see that the smallest $sigma$-algebra satisfying the required properties is the only $sigma$-algebra satisfying the required properties?

The completion theorem is first an existence theorem. The proof gives an explicit construction of the completion.

If you look into the proof, then you see that the $sigma$-algebra $Sigma^*$ constructed there is the smallest possible.
It contains the original $mathcal A$ and all subsets of sets of zero measure.

Completion of measure space is usually defined without the requirement of smallest possible $sigma$-algebra. The completion theorem gives you this smallest completion for free.

As stated the result is surely false. If $(X,mathcal A, mu)$ is already complete you are asserting that we cannot extend the measure to a larger sigma algebra. Take $mathcal A$ to be trivial sigma algebra (consisting of just the empty set and the whole space) to get obvious counter-examples.

In general, the extension you are looking for is not unique.
Let $X=[0,1]$, for example, and consider the trivial $sigma$-algebra $mathcal A=emptyset, X$. Let $mu(emptyset)=0$ and $mu(X)=1$.

This is complete, but there are a lot of different ways to extend it to a complete measure on a finer $sigma$-algebra.
For example, $mathcal B$ can be all subsets of $X$, and the measure can be any Dirac measure.

Or the Lebesgue measure is also good.