# Prove that $\,f=0$ almost everywhere.

Let $f$ be a Lebesgue integrable function on $[0,1]$ such that, for any $0 \leq a < b \leq 1$,

$$\biggl|\int^b_a f(x)\,dx\,\biggr| \leq (b-a)^2\,.$$

Prove that $f=0$ almost everywhere.

I would be thankful if somebody tell me whether my attempt is correct or not:

Consider the partition $$[0,1]=\left[0,\frac{1}{n}\right) \cup \left[\frac{1}{n},\frac{2}{n}\right) \cup \cdots \cup \left[\frac{n-1}{n},1\right].$$

I am going to use the fact that if $g$ is an integrable over a measurable set $E$, $g$ is finite almost everywhere on $E$, i.e. there is $M>0$ such that $\vert g\rvert<M$ a.e. on E, moreover $$\bigg|\int_Eg(x)dx\,\bigg|\leq M m(E)$$

Now using the hypothesis

$$\biggl|\int^{\frac{1}{n}}_{0} f(x)\,dx\biggr| \leq \biggl( \frac{1}{n} \biggr)^2 \Rightarrow |f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ \biggl[0,\frac{1}{n} \biggr)$$

$$\biggl|\int^{\frac{2}{n}}_{\frac{1}{n}} f(x)\,dx\biggr| \leq \biggl(\frac{1}{n}\biggr)^2 \Rightarrow |f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ \biggl[\frac{1}{n},\frac{2}{n}\biggr)$$

and similarly

$$\biggl|\int^{1}_{\frac{n-1}{n}} f(x)\,dx\biggr| \leq \biggl(\frac{1}{n}\biggr)^2 \Rightarrow |f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ \biggl[\frac{n-1}{n},1\biggr]$$

Hence,

$$|f(x)| \leq \frac{1}{n} \ \ a.e. \ on \ [0,1]$$ we can let $n \rightarrow \infty$ which yields that $f=0$ almost everywhere on $[0,1]$.

• Being finite almost everywhere does not imply that there is a single bound for the function almost everywhere. Consider the function $f(x) = \frac{1}{x}$ on $[0,1]$. It is finite almost everywhere yet there is not a finite $M$ such that $f(x) < M$ almost everywhere. – yoknapatawpha Dec 27 '13 at 0:14
• Look at $F(x)=\int_0^x f(t) dt$ and show that $F'(x)=0$. – abnry Dec 27 '13 at 0:16
• @yoknapatawpha Thanks ! – the8thone Dec 27 '13 at 0:18
• Do you mean Riemann integrable or Lebesgue integrable? – Matemáticos Chibchas Dec 27 '13 at 0:20
• @MatemáticosChibchas Lebesgue integrable – the8thone Dec 27 '13 at 0:21

Define $$g(x)=\int_0^x f(t)dt$$. Then $$g$$ satisfies $$|g(x)-g(y)|\leq (x-y)^2$$. This implies at once that $$g$$ is continuous, differentiable with $$g'(x)=0$$. Therefore $$g$$ is constant and $$g(0)=0$$ implies that $$g(x)=0$$.

This means that $$f$$ is integrable with $$\int_I f=0$$ for every interval $$I \subset [0,1]$$ and this extends to $$\int_B f=0$$ for every Borel measurable set $$B \subset [0,1]$$. Consider now the sets $$A_n = \{ |f| \geq \frac{1}{n}\}$$. Then the sets $$A_n$$ are Lebesgue measurable and we have $$\mu(A_n)=\sup\limits_{K \subset A_n, \text{ compact} }\mu(K)=0$$ (since compact sets are Borel measurable; $$\mu$$ is the Lebesgue measure).

Therefore $$\mu(\{f\neq 0\})=\mu(\bigcup_n A_n) =\lim_{n\to \infty} \mu(A_n)=0$$ so $$f=0$$ almost everywhere.

• Thanks ! your solution can be simplified extensively, if we use a theorem in analysis that says, if $f$ is Lebesgue integrable over $[a,b]$, and $g(x)=\int^x_a f(x)$, then $g'(x)=f(x)$ almost everywhere on $[a,b]$. Hence once you've shown that $g'(x)=0$, the result follows. – the8thone Dec 27 '13 at 1:15
• @Roozbeh-unity: I thought of that, but I think a direct solution seems more appropriate here. I think the proof of the theorem you mention is not simpler than what I did here. – Beni Bogosel Dec 27 '13 at 1:19
• I have a question, you've concluded that $g'(x)=0$ because of the following ? $$|g(x)-g(y)| < (x-y)^2 \Rightarrow \frac{|g(x)-g(y)|}{x-y} < x-y$$ and if you take the limit as $x \rightarrow y$ we ( roughly ?!) get that $g'(x)<0$ – the8thone Dec 27 '13 at 1:19
• @Roozbeh-unity: Take absolute values before dividing with $(x-y)$. – Beni Bogosel Dec 27 '13 at 1:20
• I see , Thanks a lot ! – the8thone Dec 27 '13 at 1:21

Fact. For every $\varepsilon>0$, there exists a $\delta>0$, such that $m(E)<\delta$ implies that $\int_E|f|dx<\varepsilon$.

This is due to the fact that $f$ is integrable.

We have that $$\{x: f(x)\ne 0\}=\bigcup_{k\in Z} \{x: f(x)\in (2^k,2^{k+1})\}\cup \bigcup_{k\in Z} \{x: f(x)\in (-2^{k+1},-2^{k})\}=\bigcup_{k\in Z}A^+_k\cup\bigcup_{k\in Z}A^-_k.$$ It suffices to show that $$m(A_k^+)=m(A_k^-)=0, \quad\text{for all}\,\,\, k\in \mathbb Z.$$ Assume that $m(A_k^+)=a>0$. For every $\varepsilon>0$, there exists an open set $U$, such that $A_k^+\subset U$ and $m(U\smallsetminus A_k^+)<\varepsilon$. Then $$\int_U f\,dx=\int_{A_k^+}f\,dx+\int_{U\smallsetminus A_k^+}f\,dx.$$ Clearly, $\int_{A_k^+}f\,dx\ge 2^ka$. Using the Fact above, we can choose $\varepsilon$ small enough, so that $\int_{U\smallsetminus A_k^+}|f|<2^{k-1}a$, and therefore $$\int_{U\smallsetminus A_k^+}f\,dx\ge-2^{k-1}a,$$ and hence $\int_U f\,dx\ge2^{k-1}a.$ But, as $U$ is open it can be written as a union of disjoint open intervals: $U=\cup_{n\in\mathbb N}I_n$, and $$0<\int_U f\,dx=\sum_{n\in\mathbb N}\int_{I_n}f\,dx,$$ which means that for some interval $I_n=(c,d)$ we should have $$\int_c^d f\,dx>0.$$ Using now the assumption, for every $n\in\mathbb N$, we have $$\int_c^d f\,dx=\sum_{k=1}^n\int_{c+\frac{(k-1)(d-c)}{n}}^{c+\frac{k(d-c)}{n}} f\,dx\le n\cdot \left(\frac{d-c}{n}\right)^2=\frac{(d-c)^2}{n},$$ which of course implies that $\int_c^d f\,dx=0$, which is a contradiction. Thus $f=0$ a.e.

• can you give more details on the part " we can choose epsilon small enough, so that the inequality > -2^(k-1)a is true? – DeepSea Dec 27 '13 at 1:43
• @NowOrNever: See edited version. – Yiorgos S. Smyrlis Dec 27 '13 at 8:57

The Lebesgue differentiation theorem states that if $f$ is integrable, then $f(x) = \lim_{\epsilon \downarrow 0} {1 \over 2\epsilon} \int_{x-\epsilon}^{x+\epsilon} f(t) dt$ for ae. [$m$] $x \in [0,1]$. Since $| {1 \over 2\epsilon} \int_{x-. \epsilon}^{x+\epsilon} f(t) dt | \le { (2 \epsilon)^2\over 2 \epsilon}$, we see that $f(x) = 0$ ae. [$m$] $x \in [0,1]$.