I am trying to solve this exercise:
Let $f$ be a lebesgue measurable on the line. if for any $r \in \mathbb{Q}$, $f(x+r) = f(x)$$ (a.e. x)$. Prove that there exist a constant $C$, such that $f = C (a.e.)$
I try to use the theorem that if $f$ is lebesgue integrable on $[a, b]$, then almost every point in $[a, b]$ is a lebesgue point, that is $\lim_{h_1+h_2\to 0}\frac{1}{h_1+h_2}\int_{x_0-h_1}^{x_0+h_2}|f(x)-f(x_0)|dx=0$ for $h_1\ge0, h_2\ge0, h=h_1+h_2\neq0 $ $a.e. x \in [a, b]$.
To prove that the function $f$ is integrable on every interval, I just need to prove that the function is integrable on some closed interval. So the question is: does lebesgue measurable function not lebesgue integrable on every closed interval exist?
Is this method to prove the exercise workable? Or how can I solve the exercise?
Best Answer
Yes. Define a measurable function $g:\mathbb{R} \rightarrow \mathbb{R}$ by $$g(x)=\frac{1}{x}\mathcal{X}_{(0,1]}(x).$$ Let $\{r_n\}_{n=1}^\infty$ be a countable dense subset of $\mathbb{R}$ and define $f_n(x):=\frac{1}{2^n}g(x-r_n)$. Put $f=\sup_n f_n$. Then $f$ is measurable and isn't integrable on any interval. The only issue is that it might take on the value $\infty$. To remedy this, note that for $k\in \mathbb{N}$, $$A_k:=\{x\in \mathbb{R}: f(x)> k\} \subseteq \bigcup\limits_{n=1}^\infty \{x\in \mathbb{R}: f_n(x)>k\},$$ and $\mu(\{x\in \mathbb{R}: f_n(x)>k\})=\frac{1}{k2^n}$. Hence $\mu(A_k)\leq \frac{1}{k}$. Since $A_\infty:=f^{-1}(\{\infty\})\subseteq \bigcap\limits_{k=1}^\infty A_k$ we conclude that $\mu(A_\infty)=0$. Then we can modify $f$ to be something finite, say $0$, on $A_\infty$ and the integral on any interval stays the same. The result is a finite, measurable function which isn't integrable on any interval.