Does there exist a sequence of compact rectifiable sets $C_{N}$ such that $\lim_{N\rightarrow \infty}\int_{C_{N}} x=\lambda$ for any $\lambda$?

Question

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be the function $f(x)=x$. Show that, given $\lambda \in \mathbb{R}$, there exists a sequence $C_{N}$ of compact rectifiable subsets of $\mathbb{R}$ whose union is $\mathbb{R}$, such that $C_{N}\subset Int(C_{N+1})$ for each $N$ and $$ \lim_{N\rightarrow \infty} \int_{C_{N}}f=\lambda$$. Does the integral of $f(x)$ exist?

Answer 1

Put $C_N=[-a_N+\theta_N, a_N]$, where $(a_N)$ is an increasing sequence of positive numbers tending to $\infty$ and $(\theta_N)$ is a sequence of real numbers tending to $0$, with $\theta_N<2a_N$ (so that $-a_N+\theta_N<a_N$).

We certainly have $\mathbb R=\bigcup_N C_N$; and $C_N\subset int(C_{N+1})$ provided that $\theta_{N+1}-\theta_N<a_{N+1}-a_N$.

The integral $\int_{C_N} f$ is equal to $$\frac12\left(a_N^2-(-a_N+\theta_N)^2 \right)=\frac 12\,\theta_N\times (2a_N-\theta_N) $$ Since $2a_N-\theta_N\sim 2a_N$ (because $a_N\to\infty$ and $\theta_N\to 0$), it follows that $$\left(\lim_{N\to\infty}\int_{C_N}f=\lambda\right)\iff \left(\lim_{N\to\infty} \theta_N a_N=\lambda\right)$$ So everything will be fine with $\theta_N=\lambda/a_N$, provided that $\theta_N<2a_N$ and $\theta_{N+1}-\theta_N<a_{N+1}-a_N$, i.e. $a_N^2>\lambda/2$ and $a_{N+1}-a_N>\lambda\times\left(\frac{1}{a_{N+1}}-\frac1{a_N}\right)$; which is obviously not a severe restriction on the sequence $(a_N)$.