Lebesgue Measure on (a,b) and (a,b]

Question

For an exercise I want to show that $\lambda((a,b))=\lambda((a,b])$ where I write $(a,b) = (a_1,b_1)\times(a_2,b_2)\times\dots\times(a_d,b_d)$ and $\lambda$ is the Lebesgue measure on $\mathscr{B}(\mathbb{R^d})$. I tried using a theorem that we proved earlier.

Let $A_n = (a,b_n]$ with $b_n=(b_1-\frac{1}{n},b_2-\frac{1}{n},\dots,b_d-\frac{1}n)$ so that $A_n\uparrow(a,b)$. Using the theorem we would find $$\lim\limits_{n\rightarrow\infty}\prod\limits_{k=1}^d\left(b_1-a_1-\frac{1}{n}\right)=\lim\limits_{n\rightarrow\infty}\lambda(A_n)=\lambda((a,b)).$$

This is almost equal to what we want, but we have that nasty fraction in the product. How can I justify saying the product is equal to $\lambda((a,b])$? Just for clarification, we have as definition: $$\lambda((a,b])=\lim\limits_{n\rightarrow\infty}\prod\limits_{k=1}^d(b_1-a_1).$$

Answer 1

So $(b_{k}-a_{k}-1/n)\rightarrow b_{k}-a_{k}$ as $n\rightarrow\infty$. Apply successively the product rule of limits: $(b_{1}-a_{1}-1/n)(b_{2}-a_{2}-1/n)\rightarrow(b_{1}-a_{1})(b_{2}-a_{2})$, do it in this way up to $d$, we get $(b_{1}-a_{1}-1/n)\cdots(b_{d}-a_{d}-1/n)\rightarrow(b_{1}-a_{1})\cdots(b_{d}-a_{d})$ as $n\rightarrow\infty$.

Answer 2

Assume $d>1.$ Let $\lambda_d$ be $d$-dimensional Lebesgue measure.

Notation: Let $S(j)=\{k\in \Bbb N:j\ne k\leq d\}$

For brevity let $A=(a,b] \setminus (a,b).$ We need to prove $\lambda_d(A)=0.$

We have $A=\cup_{j=1}^d A_j$ where each $A_j$ is isometric to $B_j=\{b_j\}\times \prod_{k\in S(j)}(a_k,b_k]$. So we have $$\lambda_d(A)= \lambda_d \left(\cup_{j=1}^d A_j\right)\leq \sum_{j=1}^d\lambda_d(A_j) =\sum_{j=1}^d \lambda_d(B_j).$$

Now $B_j\subset \{b_j\}\times \Bbb R^{d-1}.$ So it suffices to show that $\lambda_d(\{b\}\times\Bbb R^{d-1})=0$ for any $b\in \Bbb R.$ We do this by showing $\lambda_d(\{b\}\times \Bbb R^{d-1})\leq \epsilon$ for any $\epsilon >0.$

Given $\epsilon >0, $ for $n\in \Bbb N$ let $C_n= (b-e_n,b+e_n)\times (-n,n)^{d-1}$ where $e_n=\epsilon \cdot 2^{-n-1}\cdot (2n)^{1-d}.$ We have $ \{b\}\times \Bbb R^{d-1}\subset \cup_{n\in \Bbb N}\;C_n. $ And we have $\lambda_d(C_n)=\epsilon \cdot 2^{-n}.$

$$\text {Therefore } \quad \lambda_d \left( \{b\} \times \Bbb R^{d-1} \right) \leq \lambda_d \left( \cup_{n\in \Bbb N} \;C_n \right) \leq \sum_{n\in \Bbb N} \lambda_d(C_n)=\epsilon.$$