I'm reading a solution to the following problem.
Let $f:[0,1]\to \mathbb{R}$ be continuous with $\min_{[0,1]}f(x)=0$. Assume that for all $0\leq a<b\leq 1$,
$$\int_a^b(f(x)-\min_{[a,b]}f(y))\;dx\leq \cfrac{b-a}{2}.$$
Prove that for all $\lambda \geq 0$,
$$m\{x:f(x)>\lambda+1\}\leq \cfrac{1}{2}m \{x:f(x)>\lambda\}.$$
Here, $m$ is the Lebesgue measure. The solution proceeds by noting that the set $E_{\lambda}:=\{x:f(x)>\lambda\}$ is open and can therefore be written as a countable union of disjoint open intervals $(a_j,b_j)$.
Question: The solution then claims that $\min_{[a_j,b_j]}f(x)=\lambda$, for EACH $j$. This is the part that I don't understand. Why must $f$ attain its minimum value $\lambda$ on each of these intervals? Shouldn't it be "for some $j$"? There could be more than one such $j$, but why all $j$?
That the minimum is attained on each interval is crucial for what follows. I understand everything in the solution other than this one statement that seemingly follows from the continuity of $f$. Any clarification regarding this is appreciated. Thanks.
Link to the solution: page 9 of https://www.math.ucla.edu/~adamlott99/analysis_qual_solutions.pdf
The $(a_j,b_j)$ don't cover the whole domain, they just consist of the points where $f>\lambda$. The point then is that $f$ must equal $\lambda$ at the endpoints of these intervals, otherwise because of continuity the intervals could be expanded in a way that would keep them open. (There's a minor exception to this statement at the endpoints of the whole domain, but as they say in the solution, this doesn't affect the argument.)