If $f:\mathbb{R} \to \mathbb{C}$ is a Borel function such that $|f|=1,$ then there exist $\alpha<\beta$ such that $\int_{\alpha}^{\beta}f(x)dx \neq 0$ Let $f:\mathbb{R} \to \mathbb{C}$ be a Borel function such that $|f|=1.$
Is it true that there exist $\alpha<\beta$ such that $\int_{\alpha}^{\beta}f(x)dx \neq 0$?
Of course if $f$ was continuous then this follows immediately since there exists $\delta>0$ such that $$\left|\left|\int_{0}^\delta f(x)dx\right|-\delta\right| \leq \int_0^{\delta}|f(x)-1|dx \leq \frac{1}{2}\delta$$ so that $\int_0^{\delta}f(x)dx \neq 0.$
 A: Lebesgue differentiation theorem?
Given $a<b$, we have $\int_a^b|f| = b-a < \infty$, so $f$ is Lebesgue integrable on $[a,b]$.  For almost every point $x \in (a,b)$, we have
$$
\lim_{r \to 0^+} \frac{1}{2r}\int_{x-r}^{x+r} f = f(x).
\tag1$$
Choose such a point $x \in (a,b)$.  Then $|f(x)|=1$ so $f(x) \ne 0$.  From $(1)$ we conclude: when $r>0$ is small enough,
$$
\int_{x-r}^{x+r} f \ne 0
$$
and $a < x-r < x+r < b$.
A: Yes, moreover, for any $a < b$ there are $\alpha$, $\beta$ s.t. $a < \alpha < \beta < b$ and $\int_\alpha^\beta f(x) dx \neq 0$.
Let $A_1 = \{x \in [0, 1] | \operatorname{Re}(f(x)) > \frac{1}{2}\}$, $A_2 = \{x \in [0, 1] | \operatorname{Re}(f(x)) < -\frac{1}{2}\}$, $A_1 = \{x \in [0, 1] | \operatorname{Im}(f(x)) > \frac{1}{2}\}$, $A_4 = \{x \in [0, 1] | \operatorname{Im}(f(x)) < -\frac{1}{2}\}$.
We have $A_1 \cup A_2 \cup A_3 \cup A_4 = [0, 1]$, so, for some $k$, $\mu(A_k) > 0$. Wlog (multiplying $f$ by constant) we can assume $k = 1$.
From definition of Lebesgue measure, there is an open set set $B$ s.t. $A_1 \subseteq B$ and $\mu(B \setminus A_1) < \frac{\mu(B)}{4}$.
As open subset of $\mathbb R$, $B$ is union of at most countable many non-overlapping intervals, $B = \cup_n I_n$.
We have $$\sum_n \mu(I_n \setminus A_1) = \mu(\cup_n I_n \setminus A_1) = \mu(B \setminus A_1) < \frac{\mu(B)}{4} = \sum_n\frac{\mu(I_n)}{4}$$
As sum in left part is less than sum of right part, at least term in left part is less than corresponding term in right part, so for some $m$, $\mu(I_n \setminus A_1) < \frac{\mu(I_n)}{4}$.
Let $I_n = (\alpha, \beta)$.
Then $$\operatorname{Re}\int_\alpha^\beta f(x)\, dx = \operatorname{Re}\left(\int_{A_1 \cap I_n} f(x)\, dx + \int_{I_n \setminus A_1} f(x)\, dx\right) >
\operatorname{Re}\left(\frac{1}{2}\cdot \mu(A \cap I_n) - \mu(I_n \setminus A_n) \right) >
\operatorname{Re}\left(\frac{1}{2} \cdot \frac{3}{4}\cdot \mu(I_n) - \frac{1}{4}\cdot \mu(I_n)\right) = \frac{\mu(I_n)}{8} > 0
$$
A: Suppose not. Then for each bounded interval $I$, $f$ is integrable on $I$ and $\int_I f=0$. For a fixed $R$ and each finite union of disjoint open intervals $(I_k)_{k=1}^K$ contained in $(-R,R)$, $\int_{\bigcup_{k=1}^KI_k}f(x)dx=0$.
As an open set in $\mathbb R$ can be written as a countable at most disjoint union of open intervals, we derive that for each open subset $O$ of $(-R,R)$,
$\int_O f(x)dx=0$. By regularity of Lebesgue measure, we know that for each Borel subset $B$ of $(-R,R)$ and $\varepsilon>0$, there exists an open set $O$ such that $O\subset B$ and $\lambda(O\setminus B)<\varepsilon$.
From this, we derive that $\int_B f(x)dx=0$ hence that $f=0$ almost everywhere, which is a contradiction.
