(Uniqueness of Weak Derivatives)
A weak derivative of $u$, if exists, is uniquely defined up to a set of measure zero.
Special case proof (improvise from a book):
Let $v$ and $w$ be weak $1^{th}$-partial derivatives of $u(x,t)$ on a compact interval $U=[0,1]$, then $$ \int_{0}^{1} u \phi_{x} \: dx = - \int_{0}^{1} v \phi \: dx $$ and $$ \int_{0}^{1} u \phi_{x} \: dx = - \int_{0}^{1} w \phi \: dx $$ where $\phi$ is also in $C^{1}(U)$ with compact support in $U$. Then we must aslo have $v(x,t) = w(x,t)$.
I absolutely understood the proof but what does it mean that a weak derivative is uniquely defined up to a set of measure zero?
I found a similar question with an accepted answer here: What is set of measure zero?. But no idea the connection to the proof above.