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$A \subset [0,1]$.

$\forall_{a \in A}$ we name as $P_a$ a parabola that is tangent to $OX$ in point $(a,0)$.

$B = \bigcup_a P_a \cap [0,a] \times \mathbb{R}$

Show that: $$\lambda_2(B) = 0 \iff \lambda_1(A) = 0$$

My ideas:


We can assume the parabola to be of form: $P_a : y = k(x-a)^2$ for some constant $k$. The exact value of $k$ does not affect the measure calculations, so let's assume $k=1$ for simplicity. Then: $P_a : y = (x-a)^2$

Therefore, we have that: $$B = \bigcup_a \{ (x,y) \in [0,a] \times \mathbb{R} \ | \ y = (x-a)^2 \}$$


$$\lambda_1​(A) = 0 \implies \lambda_2​(B) = 0$$

Assume $\lambda_1(A)=0$. This means $A$ has Lebesgue measure zero. To show that $\lambda_2​(B)=0$, we need to demonstrate that the set $B$ formed by these parabolas has Lebesgue measure zero in $\mathbb{R}^2$.

  • Each parabola segment $P_a \cap [0,a]$ is contained in a very "thin" set in $R2$
  • For each $a \in A$, consider the curve $(x,(x−a)^2)$ for $x \in [0,a]$.
  • The width of this set in the $x$-direction is at most $a$.
  • The height (or $y$-value) is maximized at $x=0$ where $y=a^2$

However, since $A$ has measure zero, we can cover $A$ by intervals $\{ I_n \}$ with total length $ \Sigma \ \text{length}(I_n) < \epsilon$ for any $\epsilon > 0$. Correspondingly, the union of parabolic segments over these intervals will also have a very small area, intuitively because each $I_n$​ contributes a vanishingly small area to $B$ as $n \to \infty$.

Formally, each interval $I_n=[a_n,b_n]$ with length $b_n − a_n$ can be mapped to a vertical strip whose width is $b_n ​− a_n$​ and height proportional to $(b_n − a_n)^2$. Thus, the total area contributed by each interval is at most $(b_n − a_n) \cdot (b_n − a_n)^2 = (b_n ​− a_n​)^3$. Summing over all intervals:

$$\Sigma \ (b_n - a_n)^3 < \epsilon \ \Sigma (b_n - a_n)^2 \to \{ \text{as: } n \to \infty \implies \epsilon \to 0 \} \to 0$$

Thus, indeed: $\lambda_1​(A) = 0 \implies \lambda_2​(B) = 0$


$$\lambda_2(B) = 0 \implies \lambda_1(A) = 0$$

Assume $λ2(B)=0$. If $A$ had positive measure, then $B$ would contain a "significant" area around each $a∈A$.

  • If $A$ has positive measure, we could cover $A$ by a collection of non-overlapping intervals $\{ I_n \}$ with total length $\Sigma length(I_n) > \sigma$ for some $\sigma > 0$.
  • Each interval $I_n​ = [an_​,b_n​]$ would contribute a measurable area to $B$, since each parabolic segment corresponding to $I_n$​ has positive area proportional to the cube of the interval length.

This implies that $B$ would have positive measure, contradicting $\lambda_2(B)=0$. Thus, $A$ must have measure zero.


Both directions have been demonstrated, we have that:

$$\lambda_2(B) = 0 \iff \lambda_1(A) = 0$$

Does it look correct?

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