Integration problem, existence of a constant in a inclusion between integrable functions and essentially bounded functions Let $\mu$ be a finite positive measure on the borel sets in $\mathbb{R}$ and suppose $\mathcal{L}^{1}(\mathbb{R},\mu) \subset \mathcal{L}^{\infty}(\mathbb{R},\mu)$. 
Show that there exists $c>0$ such that if A is a Borel set with $\mu(A) > 0$ then $\mu(A) \geq c$. 
No idea at all. Can you please help?
 A: The rough idea is that if there are sets of arbitrarily small measure, then you can define unbounded integrable functions by making them get larger and larger but on subsets that are more quickly getting smaller and smaller.  This is why $L^1$ is not contained in $L^\infty$ in the case of $(0,1)$ with Lebesgue measure, for example.  
I'll try to give you a little more of the idea without giving everything away.  You can prove this with contraposition, so your assumption will be that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$.  Use this to construct an infinite sequence of disjoint Borel sets with positive measure converging to zero, say at a rate faster than $2^{-n}$.  Then define a function whose value on the $n^\text{th}$ set is $n$.
All the work is in constructing the sequence of sets.  I'd rather not spoil your fun by saying more, unless you get stuck along the way and have a specific question.

December 15 update:  To enjoy the new spoiler feature, I decided to post a sketch of one way to construct the sequence in the box below.  

 Suppose that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$.  Start with any Borel set $A_0$ such that $0\lt\mu(A_0)\lt \infty$.  (Technically the finiteness is redundant here, because $\mu$ is assumed to be finite.)  For each positive integer $n$, the hypothesis guarantees that given $A_{n-1}$ with positive measure, there is a Borel set $A_n$ such that $0\lt\mu(A_n)\lt \frac{1}{3}\mu(A_{n-1})$.  Do this for every positive integer (countable AC).  Then define a sequence of Borel sets $B_0,B_1,\ldots$ by $B_n=A_n\setminus\cup_{k\gt n}A_k$.   The inequality $\mu(A_k)\lt 3^{n-k}\mu(A_n)$ for all $k\gt n$ implies that each $B_n$ has positive measure (at least half of $\mu(A_n)$) and that $\sum_{n\geq 0}n\mu(B_n)$ converges.  If $m\gt n$, then $B_m$ is contained in $A_m$, while $B_n$ is contained in $A_n\setminus A_m$, so $B_m$ and $B_n$ are disjoint.  The function $\sum_{n\geq1}n\chi_{B_n}$ is thus in $L^1\setminus L^\infty$.

A: What Jonas Meyer did is definitely the most natural approach, but there's also a functional analysis approach to such problems I find amusing. In this case, for each $A$ with $\mu(A) > 0$ define
$$T_A(f) = {1 \over \mu(A)} \int_A f$$
 Then $T_A$ is a bounded linear map from $L^1({\mathbb R},\mu)$ to ${\mathbb C}$, and by your condition that $L^1({\mathbb R},\mu) \subset L^{\infty}({\mathbb R},\mu)$, for each $f$ there is a constant $C_f = ||f||_{\infty}$ such that  $||T_A f|| \leq C_f$ for all $A$. Hence by the uniform boundedness principle there is a single constant $C$ such that $||T_A f|| \leq C||f||_1$ for all $A$. Taking $f$ to be the characteristic function of $A$ this leads to $1 \leq C\mu(A)$, which is equivalent to what you are trying to prove.
And those of you who are familiar with the proof of the uniform boundedness principle may see the connection between this proof and Jonas's.
