Show there exists $x\in (0,1)$ such that $f(x) \leq \int_0^1 f(t) dt$ Please help me check my proof, thanks!
(a) Show there exists $x\in (0,1)$ such that 
$$f(x) \leq \int_0^1 f(t) dt.$$
Proof: when $f$ is constant a.e, the equality holds for all points except for a set of measure zero. Suppose $f$ is not constant a.e, argue by contradiction, suppose that 
$$f(x) > \int_0^1 f(t) dt $$
for $x\in (0,1)$ a.e. 
Then $\{f(x) : x\in (0,1) \text { and } f(x) > \int_0^1 f(t) dt \}$ is bounded below and there exists an $C:= \inf_{x\in (0,1) \text { and } f(x) > \int_0^1 f(t) dt } f(x)$. Now we have 
$$C \geq \int_0^1 f(t) dt$$
and 
$$\int_0^1 C dt \geq \int_0^1 f(t) dt$$
$$\int_0^1 f(t) - C dt \leq 0.$$
Observe that $f(t) - C\geq 0$ a.e. thus we must have $f(t) - C = 0$ a.e., which contradicts with the assumption that $f$ is not constant a.e. 
(b) Given an $1>\epsilon > 0$, construct a function such that the measure of the set of points that satisfy the above inequality is less than $\epsilon$.
Define $f(x) = -1$ when $x\in (0,\epsilon)$ and $f(x) = 0$ when $x\in (\epsilon, 1)$.
 A: This proof looks wrong to me. The claim that $\{ f(x): x \in (0, 1)\}$ is bounded below seems wrong. Look at, say, $f(x) = -1/x^8$ for $x \ne 0$, and $f(0) = 0$. Then that set is not bounded below. 
The place where it really messes up is when you suppose that $f(x) > \int_0^1 ...$. At this point, $x$ is a specific number in the unit interval. But in the next line, you use it as a variable in a set-specification, and things get confused. 
Edit, post-comments: 
Once the proof was modified to suppose that $f(x) > \int_0^1 f(t) dt$ for all $x \in (0, 1)$, the remaining failure is in the last couple of steps. You say that $int_0^1 f(t) −C ~dt \le 0$, but $f(t)−C\ge 0$ for all $t$, and conclude that this means $f(t)−C=0$. Unless you're assuming $f$ is continuous, this conclusion isn't justified. (Think of $C=0$, and $f(t)=0$ except for $f(0)=1$, or $f(1/2) = 1$.) 
I think that perhaps a better division would be into two cases: (1) there is a constant $C$ such that $f(t) = C$ almost everywhere, or (2) there is no such constant. The first case gets just a tiny bit trickier, but in the second case, you can conclude at the end that $f(t) = C$ almost everywhere; it thus must be true at some point of $(0, 1)$.
