$\int_I f\,dm\geq\vert I\vert$ for any interval $I$, prove $f(x)\geq 1\text{ a.e.}$ Problem:

$f$ is Lebesgue integrable function defined on $[a,b]$, and $\int_I f\,dm\geq\vert I\vert$ for any interval $I\subset [a,b]$. Prove $f(x)\geq 1 \text{ a.e. } x\in [a,b]$

As open sets can be written as the union of countable disjoint open intervals, $\int_G f\,dm\geq m(G)$ for any open set $G\subset[a,b]$. Then I try to approximate the measurable set $\{x\in [a,b]:f(x)<1\}$ with an open set and hope this would be helpful, but it does not work. I am afraid this may not be the right way to solve this problem.
 A: After you've shown that $\int_G f\, dm \ge |G|$ for all open sets $G\subset [a,b]$, use the monotone convergence theorem to show $\int_H f\, dm \ge |H|$ whenever $H$ is the countable intersection of a decreasing sequence of open sets. Next, if $A$ is a Lebesgue measurable set in $[a,b]$, there is a set $H$ as previously described containing $A$ such that $H - A$ is a null set. Use this fact to show that $\int_A f\, dm \ge |A|$.
Consider the sets $S_n := \{x\in [a,b] : f(x) < 1 - 1/n\}$ for all $n\ge 1$. Those sets increase to $S := \{x\in [a,b] : f(x) < 1\}$, so it suffices to show that $|S_n| = 0$ for all $n$. Now $$|S_n| = \int_{S_n} 1\, dm \ge \int_{S_n} \left(f + \frac{1}{n}\right)\, dm \ge |S_n| + \frac{1}{n}|S_n|$$ so that $0 \ge\frac1{n}|S_n|$, or $|S_n| = 0$.
A: 
Lebesgue's Theorem:--- Here one can use a lemma related Hardy-Littlewood maximal function
which says that for a locally Lebesgue integrable function $\varphi$ defined over
$\Bbb R^n$ we have $$\lim_{r\to
 0+}\frac{1}{m(B_r(x))}\int_{B_r(x)}\varphi(y)\,dy=\varphi(x)\text{ a.e. } x\in
\Bbb R^n.$$

See Folland's Real Analysis Theorem 3.18 for proof.
Note that according to the statement we have $\frac{\int_I f\ dm}{|I|}\geq 1$ for any interval $I\subseteq(a,b)$, so that $$f(x)=\lim_{r\to 0+}\frac{1}{m\big((x-r,x+r)\big)}\int_{(x-r,x+r)}f\ dm \geq 1\text{ for almost every }x\in (a,b).$$
