If $f =
f^+-f^-$ and $g =
g^+-g^-$ in which we have: $f^-=g^-$ and $g^+ = f^+$ almost everywhere.
Does it imply that $f=g$ almost everywhere?
I know the answer is no. but I can't find any counter example.
2026-03-29 03:18:57.1774754337
Need a counter example
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The answer is yes.
Let $A^-=\{f^-\neq g^-\}$ and $A^+=\{f^+\neq g^+\}$
Then $\mu(A^-)=\mu(A^+)=0$
By finite union, $\mu(A^- \cup A^+)=0$
And obviously $\{f\neq g\}\subset (A^- \cup A^+)$, so $f=g$ almost everywhere.