Showing $f\ge1$ a.e on $[a,b]$ Let $f$ be $\in L[a,b]$ Assume for any subinterval $I \subset [a,b]$ we have $\int_I f \geq |I|$ show that $f \geq 1$ a.e on $[a,b]$.
I started with a proof by contradiction. Assume there exists a set $E$ that has positive measure where $f < 1$ on E.  Since f  $\in L[a,b]$ we can find $|f| < N$ a.e.  I then find a union of disjoint intervals $\cup I_k$ where $m(\bigcup I_k - E) < \frac{\epsilon}{N} $ (using excision) 
I am able to use this to show that $\int_{\bigcup I_k} f < m(\bigcup I_k) + \frac{\epsilon}{N} $ but I cant seem to show $\int_{\bigcup I_k} f < m(\bigcup I_k)$ which I need for a contradiction (equality is not enough).  Help!
 A: I think this might work.  Since $f$ is integrable over $[a,b]$ for each $\epsilon >0$ there exists a $\delta >0$ where if $A \subset [a,b]$ and $m(A) < \delta$ we have $\int_A |f| < \epsilon$ .  Since $\int_E f < m(E)$ there exists a $n$ such that $\int_E f < m(E) - \frac{1}{n}$.  Let $\epsilon = \frac{1}{n} . $ Find the disjoint intervals $I_k$ such that $E \subset \cup I_k$ and $m(\cup I_k-E) < \delta$ we then have:
$\int_{\cup I_k} f = \int_E f + \int_{(\cup I_k-E)}f < m(E)-\frac{1}{n}+\epsilon = m(E) \leq m(\cup I_k)$
Since the $I_k$ are disjoint this contradicts $\int_{I_k} f >= m(I_k)$
A: Assume by contradiction that $\mathcal{L}^{1}(\{x\in\lbrack
a,b]:\,f(x)<1\})>0$. Since
$$
\{x\in(a,b):\,f(x)<1\}=\bigcup_{n=1}^{\infty}\{x\in(a,b):\,f(x)\leq1-1/n\},
$$
there exists $n$ such that the set $E=\{x\in(a,b):\,f(x)\leq1-1/n\}$ has
measure $\mathcal{L}^{1}(E)=M>0$. Since $f$ is integrable, for every
$\varepsilon>0$ there exists $\delta>0$ such that if $F\subseteq\lbrack a,b]$
is a measurable set with $\mathcal{L}^{1}(F)\leq\delta$, then $\int
_{F}|f(x)|\,dx\leq\varepsilon$ (I will not prove this. It is a standard
property of integrable functions). Fix $\varepsilon<M/n$ and an open set
$U\subseteq(a,b)$ which contains $E$ and such that $\mathcal{L}^{1}(U\setminus
E)\leq\delta$. Then $U$ can be written as union of countably many disjoint
intervals $I_{k}$. It follows that
$$
\int_{U}f(x)\,dx=\sum_{k}\int_{I_{k}}f(x)\,dx\geq\sum_{k}\mathcal{L}^{1}%
(I_{k})=\mathcal{L}^{1}(U)\geq M.
$$
On the other hand,
\begin{align*}
\int_{U}f(x)\,dx  & \leq\int_{U\setminus E}f(x)\,dx+\int_{E}f(x)\,dx\leq
\varepsilon+(1-1/n)\mathcal{L}^{1}(E)\\
& \leq\varepsilon+(1-1/n)M.
\end{align*}
Thus, $M\leq\varepsilon+(1-1/n)M$, which gives $M/n\leq\varepsilon$, a
contradiction. 
There is a much simpler proof using Lebesgue points, that is, the fact that for
$\mathcal{L}^{1}$ a.e. $x\in(a,b)$ there exists
$$
\lim_{r\rightarrow0^{+}}\frac{1}{2r}\int_{x-r}^{x+r}f(t)\,dt=f(x),
$$
but probably it is too advanced.
