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We all know that if $f\leq{}g$ in $[a,b]$ then $$ \int_a^bf\,dx\leq\int_a^bg\,dx $$ now, imagine that we have $f<g$, is it true that

$$ \int_a^bf\,dx<\int_a^bg\,dx $$

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  • $\begingroup$ Hint: make it one integral in your inequality. $\endgroup$ – Steve D May 22 '16 at 1:38
  • $\begingroup$ @SteveD's hint : this reduces to the question of whether or not $h >0 \implies \int h > 0$ $\endgroup$ – Chill2Macht May 22 '16 at 1:39
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    $\begingroup$ This is indeed true, but I am not sure if we have any proof that avoids ideas from measure theory. $\endgroup$ – Sangchul Lee May 22 '16 at 1:46
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    $\begingroup$ @SangchulLee: huh? $(g-f)$ has a positive minimum ($[a,b]$ is compact)... $\endgroup$ – Steve D May 22 '16 at 2:10
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    $\begingroup$ @SteveD If it's not continuous, the infimums may be $0$. $\endgroup$ – MathematicsStudent1122 May 22 '16 at 2:11
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Here is an argument which borrowed ideas from measure theory, but does not assume any direct knowledge on it.


Proof. It suffices to prove the following claim:

Claim. If $h \in \mathscr{R}([a, b])$ satisfies $h \geq 0$ and $\int_{a}^{b} h \, \mathrm{d}x = 0$, then $h(c) = 0$ for some $c \in [a, b]$.

Step 1. To this end, assume that $h$ satisfies the assumptions of the claim. Then we have the following observation:

Observation. For any $\epsilon > 0$ and $\delta > 0$, there exists a relatively open subset $U \subseteq [a, b]$ such that

  1. $U$ is the union of finitely many relatively open subintervals of $[a, b]$,
  2. the lengths of $U$ is less than $\delta$, and
  3. $\{ x \in [a, b] : h(x) > \epsilon \} \subseteq U$.

We first check that this indeed implies the claim. For each $n \geq 1$, choose $U_n$ as in Observation with $\epsilon = 1/n$ and $\delta = 3^{-n}(b-a)$, so that

  1. the length of $U_n$ is less than $3^{-n}(b-a)$, and
  2. $\{ x \in [a, b] : h(x) > 1/n \} \subseteq U_n$.

Then we find that

$$ \{ x \in [a, b] : h(x) > 0 \} = \bigcup_{n=1}^{\infty} \{ x \in [a, b] : h(x) > 1/n \} \subseteq \bigcup_{n=1}^{\infty} U_n. $$

Now assume otherwise that $h > 0 $ on all of $[a, b]$. Then it follows that $\bigcap_{n=1}^{\infty} U_n = [a, b]$ and thus $\{ U_n : n \geq 1 \}$ is an open cover of $[a, b]$. So we can pick a finite subcover, say $\{ U_{n_1}, \dots, U_{n_K} \}$. This implies that

$$ [a, b] = U_{n_1} \cup \cdots \cup U_{n_K}. $$

This is a contradiction since the right-hand side has length at most

$$\sum_{n=1}^{\infty} 3^{-n}(b-a) < b-a. $$

Step 2. It now remains to prove the observation. (The proof is essentially a variant of the Markov's inequality.)

Choose a partition $P$ such that $U(P, h) < \delta \epsilon$. Write $P = \{a = x_0 < \cdots < x_N = b\}$ an define $M_j = \sup_{[x_{j-1}, x_j]} h$ and $\Delta x_j = x_j - x_{j-1}$. Then we know that $U(P, h) = \sum_{j=1}^{N} M_j \Delta x_j < \delta \epsilon$. On the other hand, let $J$ be the set of indices $j$ for which $M_j > \epsilon$. Then

$$ \sum_{j \in J} \Delta x_j \leq \frac{1}{\epsilon} \sum_{j \in J} M_j \Delta x_j \leq \frac{1}{\epsilon} U(P, h) < \delta $$

and that $\cup_{j \notin J} [x_{j-1}, x_j]$ is a finite union of closed intervals on which $h \leq \max_{j \notin J} M_j \leq \epsilon$ holds. Therefore the observation follows by taking $U$ as the complement of $\cup_{j \notin J} [x_{j-1}, x_j]$.

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Suppose on the contrary that $\int f=\int g$, we can assume $f,g$ vanishes outside the interval, then they are both $\mathscr L^1$, then we have $$\int_\Bbb R(f-g)dx=0. $$ since $f- g\ge 0$, indeed $f=g$ a.e. on $\Bbb R$, contradiction.

I'd appreciate an elementary proof without any appeal to Lebesgue theory.

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    $\begingroup$ It suffices to show that there is at least one point where $f-g$ is continuous. This answer gives a proof which does not formally require measure theory, although it has a measure-theoretic flavor. It would be nice to see a more elementary proof, but i haven't found one. $\endgroup$ – Bungo May 22 '16 at 6:04
  • $\begingroup$ @Bungo fair observation. Reduces the problem to a considerably more trivial one. $\endgroup$ – Vim May 22 '16 at 6:25
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Assuming $$f+h=g$$

and $$h>0$$

$$\int_a^b f dx+\int_a^b h dx=\int_a^b g dx$$

since

$$\int_a^b h dx>(b-a)\times\min(h(x))>0$$ We can write

$$\int_a^b f dx < \int_a^b g dx$$

but what if $h$ has no minimum? it is enough to find any piece where $h$ has a minimum to prove $\int_a^b h dx >0$. Unless $h$ has no minimum at any neighborhood. But if $f$ and $g$ are integrable then $h$ must be integrable too.

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  • $\begingroup$ Might be quite tricky to show that in at least one neighbourhood $h$ attains a minimum. The idea is inspiring, though. $\endgroup$ – Vim May 22 '16 at 3:25
  • $\begingroup$ @Vim Not that tricky. Riemann integrable functions are continuous almost everywhere. $\endgroup$ – MathematicsStudent1122 May 22 '16 at 3:25
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    $\begingroup$ @MathematicsStudent1122 that's an appeal to measure theory, which I think Arashium wants to avoid using here. $\endgroup$ – Vim May 22 '16 at 3:28
  • $\begingroup$ @Vim, Arashium is an engineer and not a mathematician. I have no idea about measure theory :) $\endgroup$ – Arashium May 22 '16 at 3:31
  • $\begingroup$ @Vim Can we prove the weaker statement $\text{discontinuous everywhere} \Longrightarrow \text{not Riemann integrable}$ without measure theory? If so, we're done. Intuitively, it seems clear; in such a situation, regardless of how fine one makes a partition, the $\sup$ and $\inf$ of $f$ in every subinterval will still be "far away" from each other. $\endgroup$ – MathematicsStudent1122 May 22 '16 at 5:38

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