How exactly does proof by contradiction work in this circumstance? (Proving property of Lebesgue Integral)
Solution 1:
The negation of "for every $\varepsilon$, (something is true about $\varepsilon$)" is not "for every $\varepsilon$, (something is false about $\varepsilon$)" as you have written, but rather "there exists $\varepsilon$ such that (something is false about $\varepsilon$)".