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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):

  1. Show $\lambda_2(B$ x $\{a\})$ and $\lambda_2(\{a\}$ x $B)=0$, $a\in\mathbb{R},B\in\mathcal{B}(\mathbb{R})$.

  2. 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.

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3 Answers 3

up vote 3 down vote accepted

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

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Think about the open sets which form a basis for the Borel subsets of $\mathbb{R}$ and first try to prove (1) for these sets.

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For 1, note that $B\times{a}=\cup_n (B\cap[-n,n])\times{a}$, and that $\lambda_2([-n,n]\times{a})\leq\lambda_2([-n,n]\times(a-1/n^2,a+1/n^2))=4/n$.

For 2, just note that the upper and lower Riemmann sums give you sets of rectangles that contain and are contained in your set respectively, and thus the measure of the set has to agree with the integral.

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