Lebesgue measure on Riemann integrable function in $\mathbb{R}^2$

measure-theory

As mentioned in a few of my other questions, I am new to measure theory and learning it on my own. I came across an interesting exercise and I would be grateful for/interested in any thoughts.

Setup:

Let $\lambda_2$ be a Lebesgue measure on $\mathbb{R}^2$ such that: $\lambda_2((a,b]$x$(c,d])=(b-a)(d-c)$ for all finite, real $a < b, c < d$.

Questions (The exercise mentions parenthetically that these questions can he proven without referring to any results or information beyond what is given):

Show $\lambda_2(B$ x $\{a\})$ and $\lambda_2(\{a\}$ x $B)=0$, $a\in\mathbb{R},B\in\mathcal{B}(\mathbb{R})$.
For $f(x) \geq 0$ Riemann-integrable on the finite interval [a,b], show that

$$\lambda_2(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\})=\int_a^b f(x)dx.$$

For 2., what little I have grasped with measure theory is that in $\mathbb{R}^2$, measure works very similarly to "area", but what is tripping me up is the caveat that these assumptions/results are not needed.

Any help is greatly appreciated as always.

Solution 1:

Let $\lambda_2^\ast$ be the Lebesgue outer measure on $\Bbb R^2$.

1. Let $n\in\Bbb Z$. Note that for any $\epsilon\gt 0$ $$\{b\}\times (n,n+1]\subset (b-\epsilon,b]\times (n,n+1]$$ so that $$\lambda_2^\ast(\{b\}\times (n,n+1])\leq \lambda_2^\ast((b-\epsilon,b]\times (n,n+1])=\epsilon.$$ Since $\epsilon$ is arbitrary, this proves $\lambda_2^\ast(\{b\}\times (n,n+1])=0$ and therefore by the definition of the Lebesgue measure $\{b\}\times (n,n+1]$ is measurable and $$\lambda_2(\{b\}\times (n,n+1])=0,$$ for each $n\in\Bbb Z$. Therefore the set $$\{b\}\times\Bbb R=\bigcup_{n\in\Bbb Z} \{b\}\times (n,n+1]\tag{1}$$ has measure $0$.

Consider $E\subseteq\Bbb R$ an arbitrary set of real numbers. Since $$\{b\}\times E\subseteq \{b\}\times\Bbb R$$ in view of $(1)$ we conclude that $$\lambda_2(\{b\}\times E)=0.$$ The other part is very similar.

2. Consider $P_n=\{x_0,\ldots,x_n\}$ the partition of $[a,b]$ given by $$x_0=a,\quad x_k=\frac{k}{n}(b-a),\ \text{ for } k\in\{1,\ldots,n\}.$$ For each $k\in\{1,\ldots,n\}$ define $$m_k=\inf f([x_{k-1},x_k])\qquad M_k=\sup f([x_{k-1},x_k]).$$ Note that $$[x_{k-1},x_k]\times [0,M_k]=\{x_{k-1}\}\times]0,M_k]\cup]x_{k-1},x_k]\times\{0\}\cup ]x_{k-1},x_k]\times ]0,M_k]\tag{3}$$ since this union is disjoint, by $1.$ we get $$\lambda_2([x_{k-1},x_k]\times [0,M_k])=\lambda_2(]x_{k-1},x_k]\times ]0,M_k])=\frac{k}{n}(b-a)\cdot M_k\quad \forall k\in\{1,\ldots,n\}. \tag{2}$$ For each $n\in\Bbb N$ define $$L_n=\frac{b-a}{n}\sum_{k=1}^n m_k\qquad U_n=\frac{b-a}{n}\sum_{k=1}^n M_k.$$ Provided that $f$ is Riemann integrable on $[a,b]$ we have $$\lim_{n\to\infty} L_n=\int_a^b f=\lim_{n\to\infty} U_n.$$ Now, notice that, for each $n\in\Bbb N$ $$\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}\subseteq Z_n\cup \bigcup_{k=1}^n ]x_{k-1},x_k]\times ]0,M_k],$$ where, in the light of $(3)$, $\lambda_2(Z_n)=0$. So $$\lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\})\leq \lambda_2^\ast\left(Z_n\cup \bigcup_{k=1}^n ]x_{k-1},x_k]\times ]0,M_k]\right)\leq U_n,$$ letting $n\to\infty$ we obtain $$\lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\})\leq \int_a^b f\tag{4}$$

Now, fix $n\in\Bbb N$. Note that for any covering $\{]a_k,b_k]\times]c_k,d_k]\}_{k\in\Bbb N}$ we have $$\bigcup_{k=1}^n [x_{k-1},x_k]\times[0,m_k]\subseteq \{(x,y):0\leq y\leq f(x),a\leq x\leq b\}\subset \bigcup_{k\in\Bbb N} ]a_k,b_k]\times]c_k,d_k]$$ then by similar arguments to the given above we get $$L_n\leq \sum_{k=1}^\infty (b_k-a_k)(d_k-c_k)$$ so $$L_n\leq \lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}).$$ Since this is true for each $n\in\Bbb N$, by letting $n\to\infty$ we get $$\int_a^b f\leq \lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}).$$ This joined with$(4)$ says $$\int_a^b f=\lambda_2^\ast(\{(x,y):0\leq y\leq f(x),a\leq x\leq b\}).$$