# $A \subset [0,1]$. $P_a$ is a parabola tangent to OX in $(a,0)$. $B= \bigcup_a P_a \cap [0,a] \times R$. Show: $\lambda_2(B)=0 \iff \lambda_1(A)=0$

$$A \subset [0,1]$$.

$$\forall_{a \in A}$$ 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?