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Let consider the function $f:\mathbb R \rightarrow I$, where $I$ is $(0,1)$. Consider that $f$ is both continuous and bijective. Let's assume that it is strictly increasing. Now consider a real number $\alpha \in I$.

What do we have to prove is that for every $\alpha \in I$, there are the irrational numbers $a$ and $b$ such that:

$$\int_{a}^{a+1} f(x) \,dx\ = \alpha+\int_{b}^{b+1} f(x) \,dx\ $$

In other words, that there are two irrational numbers $a $ and $b$ such that the area under the function $f$ between the points $a$ and $1+a$ is greater than the area under the function $f$ between the points $b$ and $1+b$ with the value $\alpha$.

$\textbf{My attempt.}$

What I got is:

Since $f$ is both continuous and bijective then it is strictly monotonous. The Hypothesis told us that it is strictly increasing. So we are fine at this point.

What we have to prove is that the area under the function $f$ between the points $a$ and $1+a$ is equal with $\alpha$ plus the area under the function $f$ between the points $b$ and $1+b$.

This is what seems to be the area under function $f$.

enter image description here

What do we have to prove is that for every $\alpha \in I$, there are the irrational numbers $a$ and $b$ such that:

$$\int_{a}^{a+1} f(x) \,dx\ = \alpha+\int_{b}^{b+1} f(x) \,dx\ $$

shangq_tou
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  • What is $b$? It seems to be introduced abruptly – spaceman Nov 01 '21 at 14:20
  • It may help to first rewrite the problem as: Prove that for any $\alpha \in I = (0,1)$, there exists irrationals $a,b\in\mathbb{R}\backslash\mathbb{Q}$ such that $$\alpha = \int_{0}^{1}f(x+a) - f(x+b) dx.$$ Now, for any $s\in\mathbb{R}$, letting $$ F(s) := \int_{0}^{1} f(x + s) dx, $$ we get that $F'(s) = f(1 + s) - f(s)$ for any $s\in\mathbb{R}$. One could possibly take it from here. – spaceman Nov 01 '21 at 15:37
  • @spaceman May you explain to me how you continued from there? – shangq_tou Nov 02 '21 at 22:38

2 Answers2

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Following from spaceman's comment, let $$F(s)=\int_0^1f(x+s)dx.$$ Then $F'(s)=f(s+1)-f(s)>0$, since $f$ is strictly increasing. Further, we also have the bounds $$f(s+1)\geq F(s)\geq f(s)$$ We conclude that $F(x)$ is also a strictly increasing continuous bijection $\mathbb{R}\rightarrow I$. Now let $\alpha\in I$. Then for each $b\in\mathbb{R}$ with $F(b)<1-\alpha$, there is a unique $a\in\mathbb{R}$ such that $F(a)-F(b)=\alpha$ given by $a=F^{-1}(F(b)+\alpha)$. There are uncountably many choices for $b\in \mathbb{R}-\mathbb{Q}$, and only countably many choices for $a\in\mathbb{Q}$, so there must be some choice of $a$ and $b$ both irrational.

Jacob
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@winic92046 This is too long for a comment, so have written it here. But know that this is just a comment for now.

A train of thought:

First, given any real number $\alpha\in(0,1)$, it is possible to show that there exist two irrationals $\beta, \gamma$ such that $\alpha = \beta + \gamma$.

@winic92046 A train of thought:

First, given any real number $\alpha\in(0,1)$, [there exist]](Prove that any real number can be expressed as the sum of two irrational numbers) two irrationals $\beta, \gamma$ such that $\alpha = \beta + \gamma$.

Now, if we can show that $F\vert_{\mathbb{R}\backslash\mathbb{Q}} : \mathbb{R}\backslash\mathbb{Q} \to \mathbb{R} $ is surjective on some semi-infinite space of irrationals (or reals), we could continue by working backwards.

In essence, we could continue by showing that for any irrational $\beta$, there exists an irrational $a$ such that $\beta = F(a)$. Similarly, for $\gamma$, there exists an irrational $b$ such that $\gamma = F(b)$. From here, we would then be able to conclude the claim.

I don't imagine it will be quite as simple as this, as there are some details that I neglected, but I believe a similar approach is worth trying.

spaceman
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