$A \subset [0,1]$.
$\forall_{a \in A}$ 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?