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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 $x\in (\epsilon, 1)$.

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  • $\begingroup$ You are assuming $f$ is continuous? $\endgroup$ – Prahlad Vaidyanathan Jul 21 '14 at 11:59
  • $\begingroup$ @PrahladVaidyanathan no, $f$ is only integrable. $\endgroup$ – Xiao Jul 21 '14 at 12:00
  • $\begingroup$ Integrable and $-\infty < \int_0^1 f(t)dt < \infty$ I assume?! $\endgroup$ – Winther Jul 21 '14 at 12:01
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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 $(0, 1)$.

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  • $\begingroup$ Edited, my assumption is $f(x) > \int_0^1 f(t) dt$ for all $x\in (0,1)$ $\endgroup$ – Xiao Jul 21 '14 at 11:59
  • $\begingroup$ Thanks. That's better. Now the failure is in the last couple of steps. You say that $\int f-C \le 0$, but $f-C \ge 0$, and conclude that this means $f - C = 0$. Unless you're assuming $f$ is continuous, this conclusion isn't justified. (Think of $C = 0$, and $f = 0$ except that $f(0) = 1$.) $\endgroup$ – John Hughes Jul 21 '14 at 12:18
  • $\begingroup$ See additional post-comment notes in my answer. $\endgroup$ – John Hughes Jul 21 '14 at 12:59
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    $\begingroup$ Here's an improved proof, by contradiction. Suppose that for all $x \in (0, 1)$, $f(x) > C = \int_0^1 f(t)~ dt$. Then $\int_0^1 f(x)~dx > \int_0^1 C ~ dx = C$. That's a contradiction. $\endgroup$ – John Hughes Jul 22 '14 at 2:56
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    $\begingroup$ @6005: Agreed. Since it has already caused a chain of actions on MSE, I won't delete that comment anymore, though. $\endgroup$ – Alex M. Aug 20 '16 at 6:40

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