MHB Proof of Outer Lebesgue Measure Monotonicity and Subadditivity

[mathmari]

Hey!

I am looking at the proof that the outer Lebesgue measure that is defined by $$m^*(A)=inf\{\sum_{n=1}^{\infty}v(R_n):A \subset\cup_{n=1}^{\infty}R_n , ( \text{ where } R_n \text{ are open rectangles}) \}$$ is actually an outer measure.

($v(R_n)$, the volume of R is the product of the lengths $I_j$.)

To show that it is an outer measure, we have to show the monotonicity and the subadditivity.

For the subadditivity:

If $A_n \subset \mathbb{R}$ then $$m^* \left ( \cup_{n=1}^{\infty} A_n\right ) \leq \sum_{n=1}^{\infty}m^*(A_n)$$

We suppose that $m^*(A_n)<+\infty, \forall n$.

Let $\epsilon>0$. Then for each $n$ there are rectangles $R_j^n,j=1,2,...$ such that $A_n \subset \cup_{j=1}^{\infty}R_j^n$ and $\sum_{j=1}^{\infty}v(R_j^n)<m^*(A_n)+\frac{\epsilon}{2^n}$.

Then $\cup_{n=1}^{\infty}A_n\subset\cup_{j,n}R_j^n$

$$\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}v(R_j^n)<\sum_{n=1}^{\infty}m^*(A_n)+\epsilon$$

$$m^* \left ( \cup_{n=1}^{\infty}A_n \right ) \leq \sum_{j,n}v(R_j^n)<\sum_{n=1}^{\infty}m^*(A_n)+\epsilon \\ \Rightarrow m^* \left ( \cup_{n=1}^{\infty}A_n\right ) \leq \sum_{n=1}^{\infty}m^*(A_n)$$Could you explain me this proof??

 

[Euge]

Hey!

I am looking at the proof that the outer Lebesgue measure that is defined by $$m^*(A)=inf\{\sum_{n=1}^{\infty}v(R_n):A \subset\cup_{n=1}^{\infty}R_n , ( \text{ where } R_n \text{ are open rectangles}) \}$$ is actually an outer measure.

($v(R_n)$, the volume of R is the product of the lengths $I_j$.)

To show that it is an outer measure, we have to show the monotonicity and the subadditivity.

For the subadditivity:

If $A_n \subset \mathbb{R}$ then $$m^* \left ( \cup_{n=1}^{\infty} A_n\right ) \leq \sum_{n=1}^{\infty}m^*(A_n)$$

We suppose that $m^*(A_n)<+\infty, \forall n$.

Let $\epsilon>0$. Then for each $n$ there are rectangles $R_j^n,j=1,2,...$ such that $A_n \subset \cup_{j=1}^{\infty}R_j^n$ and $\sum_{j=1}^{\infty}v(R_j^n)<m^*(A_n)+\frac{\epsilon}{2^n}$.

Then $\cup_{n=1}^{\infty}A_n\subset\cup_{j,n}R_j^n$

$$\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}v(R_j^n)<\sum_{n=1}^{\infty}m^*(A_n)+\epsilon$$

$$m^* \left ( \cup_{n=1}^{\infty}A_n \right ) \leq \sum_{j,n}v(R_j^n)<\sum_{n=1}^{\infty}m^*(A_n)+\epsilon \\ \Rightarrow m^* \left ( \cup_{n=1}^{\infty}A_n\right ) \leq \sum_{n=1}^{\infty}m^*(A_n)$$Could you explain me this proof??

Hi mathmari,

The proof displayed is incomplete. Everything is fine until the last three lines of inequalities, where there is no explanation. There also are two cases to consider -- one which deals with $m^*(A_k) = +\infty$ for some $k$, and the other which you've displayed. If $m^*(A_k) = +\infty$ for some $k$, then suppose $\{R_j\}$ is an arbitrary covering of $\cup A_n$ by open rectangles. These rectangles also cover $A_k$, and thus $\sum_{j = 1}^\infty v(R_j) = +\infty$. This shows that $m^*(\cup_{n=1}^\infty A_n) = +\infty$.

Now let's go back to your case, where $m^*(A_n) < +\infty$ for all $n$. We want to show that $m^*(\cup_{n=1}^\infty A_n) \le \sum_{n=1}^\infty m^*(A_n)$, which is equivalent to showing

$$ m^*(\cup_{n=1}^\infty A_n) \le \sum_{n=1}^\infty m^*(A_n) + \epsilon$$ for all $$ \epsilon > 0.$$

By definition of the infimum of a set $S \subset \Bbb R$, any number greater than $\inf S$ is not a lower bound for $S$. Keeping this in mind, we know that for every $\epsilon > 0$ and $n \in \Bbb N$, $m^*(A_n) + \frac{\epsilon}{2^n}$ is not a lower bound for the set

$$S_n := \{\sum_{j = 1}^\infty v(R_{j,n})\, |\, \text{$R_{j,n}$ are open rectangles such that $A_n \subset \cup_{j=1}^\infty R_{j,n}$}\}.$$

Therefore, for each $n$, there exists a covering $\{R_{j,n}\}_{j= 1}^\infty$ of $A_n$ by open rectangles such that

$$ \sum_{j=1}^\infty v(R_{j,n}) < m^*(A_n) + \frac{\epsilon}{2^n}.$$

Since $\cup_{j=1}^\infty R_{j,n} \supset A_n$ for all $n$, we have $\cup_{j,n} R_{j,n} \supset \cup_{n = 1}^\infty A_n$. Hence, by definition of $m^*(\cup_{n = 1}^\infty A_n)$ and the previous inequality,

$$m^*(\cup_{n=1}^\infty A_n) \le \sum_{j,n} v(R_{j,n}) = \sum_{n = 1}^\infty \sum_{j = 1}^\infty m^*(A_n) \le \sum_{n = 1}^\infty \left(m^*(A_n) + \frac{\epsilon}{2^n}\right) = \sum_{n = 1}^\infty m^*(A_n) + \epsilon.$$

Since $\epsilon$ was arbitrary, the result follows.

 

[mathmari]

(Happy)(Happy)(Happy)(Happy)

Hi mathmari,

The proof displayed is incomplete. Everything is fine until the last three lines of inequalities, where there is no explanation. There also are two cases to consider -- one which deals with $m^*(A_k) = +\infty$ for some $k$, and the other which you've displayed. If $m^*(A_k) = +\infty$ for some $k$, then suppose $\{R_j\}$ is an arbitrary covering of $\cup A_n$ by open rectangles. These rectangles also cover $A_k$, and thus $\sum_{j = 1}^\infty v(R_j) = +\infty$. This shows that $m^*(\cup_{n=1}^\infty A_n) = +\infty$.

Now let's go back to your case, where $m^*(A_n) < +\infty$ for all $n$. We want to show that $m^*(\cup_{n=1}^\infty A_n) \le \sum_{n=1}^\infty m^*(A_n)$, which is equivalent to showing

$$ m^*(\cup_{n=1}^\infty A_n) \le \sum_{n=1}^\infty m^*(A_n) + \epsilon$$ for all $$ \epsilon > 0.$$

By definition of the infimum of a set $S \subset \Bbb R$, any number greater than $\inf S$ is not a lower bound for $S$. Keeping this in mind, we know that for every $\epsilon > 0$ and $n \in \Bbb N$, $m^*(A_n) + \frac{\epsilon}{2^n}$ is not a lower bound for the set

$$S_n := \{\sum_{j = 1}^\infty v(R_{j,n})\, |\, \text{$R_{j,n}$ are open rectangles such that $A_n \subset \cup_{j=1}^\infty R_{j,n}$}\}.$$

Therefore, for each $n$, there exists a covering $\{R_{j,n}\}_{j= 1}^\infty$ of $A_n$ by open rectangles such that

$$ \sum_{j=1}^\infty v(R_{j,n}) < m^*(A_n) + \frac{\epsilon}{2^n}.$$

Since $\cup_{j=1}^\infty R_{j,n} \supset A_n$ for all $n$, we have $\cup_{j,n} R_{j,n} \supset \cup_{n = 1}^\infty A_n$. Hence, by definition of $m^*(\cup_{n = 1}^\infty A_n)$ and the previous inequality,

$$m^*(\cup_{n=1}^\infty A_n) \le \sum_{j,n} v(R_{j,n}) = \sum_{n = 1}^\infty \sum_{j = 1}^\infty m^*(A_n) \le \sum_{n = 1}^\infty \left(m^*(A_n) + \frac{\epsilon}{2^n}\right) = \sum_{n = 1}^\infty m^*(A_n) + \epsilon.$$

Since $\epsilon$ was arbitrary, the result follows.

I understand! Thank you very much for the explanation! (Happy)