Lower bound on function given lower bound on integral If we are given a continuous function $f : [0,1]\to[0,1] $ and $\int_0^1 f dx > \varepsilon$ then can we put a lower bound on the function on some finite union of disjoint intervals such that the sum of their lengths and the bound on the function depend solely on $\varepsilon $. That is, can we find $\delta $ and $a $ as functions of $\varepsilon $ such that there exists a set of $n$ intervals, $\sum_1^n b_i-a_i > \delta $ and $f>a $ at every point of these intervals.
 A: Here is an argument using a little bit of measure theory and Lebesgue integration theory. Let $A = \{ x \in [0,1] : f(x) \le \epsilon / 2 \}$ and $B$ its complement in $[0,1]$. Then $\int_A f \le \epsilon / 2$, so $\int_B f > \epsilon / 2$. On the other hand, since $f$ is bounded above by $1$, you get $\int_B f \le \lambda(B)$ where $\lambda$ is Lebesgue (length) measure, so $\lambda (B) > \epsilon / 2$. Since $B$ is a relatively open subset of $[0,1]$, it is a countable union of intervals, so you can find finitely many intervals in $B$ with total length $\ge \epsilon / 2$. So you get the result with $a = \delta = \epsilon / 2$.
A: Generalization: Suppose $f:[0,1]\to [0,1]$ is Riemann integrable and $\int_0^1 f >\epsilon.$ Then there is a collection of pairwise disjoint open subintervals of $[0,1]$ of equal length, the sum of whose lengths is $> \epsilon/2,$ and such  that $f>\epsilon/2$ on each subinterval.
Proof: For any $n\in \mathbb N$ we consider the uniform partition of $[0,1]$ into $n$ subintervals $I_k$ of length $1/n.$ Since $f$ is Riemann integrable we can choose and fix an $n$ large enough so that
$$\sum_{k=1}^{n}m_k\cdot\frac{1}{n} > \epsilon,$$
where $m_k = \inf_{I_k}f.$ Let $A= \{k:m_k>\epsilon/2\},B= \{k:m_k\le\epsilon/2\}.$ Then
$$\sum_{k\in A}1\cdot\frac{1}{n} + \sum_{k\in B}(\epsilon/2)\cdot\frac{1}{n} >\epsilon.$$
This implies (throwing away a lot) that
$$|A|\frac{1}{n} + n(\epsilon/2)\frac{1}{n} > \epsilon \implies |A|\frac{1}{n} > \epsilon/2.$$
(Here $|\,\,|$ denotes cardinality.) Thus the sum of the lengths of the subintervals $I_k$ corresponding to $k\in A$ is $> \epsilon/2,$ and since $f >\epsilon/2$ on each such subinterval, we're done. (Take the interiors of these subintervals if you want them disjoint.)
