Rigour is lost in the real world. But that does not mean I shouldn’t continue to pursue and practise my mathematical skills. Just like a programmer might learn to code by writing programs, to improve my mathematical thinking I must read other people’s proofs and write my own.

(emphasis mine) Adapted from “Why there is no Hitchhiker’s Guide to Mathematics for Programmers”.

So perhaps here will be a collection of (mostly other people’s) proofs, until I feel confident writing my own. After all, pure mathematics was by far my worse course in university studies…

There will be plenty of mistakes, but regardless. Let us begin.

email me if you spot something wrong!

#Measure Theory

Where else to begin than with my worse subject in university? Real Analysis.

In a subject that essentially everything was rote learnt, and with foundations (sort of, not really in a practical learning sense) in probability theory, this will be my (not necessarily ideal) starting point.

Goal of today: Prove Monotonicity, Subaddivity, Infinite Unions, Infinite Intersections under measure theory. Hopefully later move onto Ham Sandwich Theorem.

\( \sigma\)-algebra (Definition)

Let \(F\) be a collection of subsets of a sample space \(\Omega\).

Then \(F\) is a \(\sigma\)-algebra if and only if :

  1. The empty set is in \(F\).
  2. If \(A \in F\) then the complement \(A^c \in F\).
  3. If \(A_i \in F\), \(i=1,2,…\), then their union \(\cup A_i \in F\)

Measure (Definition)

Let \((\Omega, F)\) be a measurable space. A set function \(\nu\) defined on \(F\) is called a measure if and only if :

  1. \(0 \le \nu(A) \le \infty\) for any \(A\in F\)
  2. \(\nu(\emptyset)=0\)
  3. If \(A_i \in F\), \(i=1,2,…\), and \(A_i\)’s are disjoint for any \(i \neq j\), then $$ \nu (\cup_{i=1}^\infty A_i) = \sum_{i=1}^{\infty} \nu (A_i)$$

Monotonicity, Subaddivity, Infinite Unions, Infinite Intersections (Proof)

Let \((\Omega, F, \nu)\) be a measurable space.

1. Monotonicity

If \(A \subset B\), then \(\nu(A)\le \nu(B)\).

Proof

Since,
$$A \subset B$$ $$B = A \cup (A^c \cap B)$$ $$ A \text{ and } A^c \cap B \text{ are disjoint}$$

By the definition of a measure (2), (since they are disjoint) $$\nu(B) = \nu(A) + \nu(A^c \cap B)$$

But \(0\ge\nu(A^c \cap B)\) from measure (1) so then it follows that $$\nu(A)\le \nu(B)$$ as required.

2. Subaddivity

For any sequence \(A_1, A_2,…\), $$ \nu(\cup_{i=1}^\infty A_i) \le \sum_{i=1}^\infty \nu(A_i)$$

Proof $$ \nu(A\cup B) = \nu(A) + \nu(B) - \nu(A \cap B)$$

By measure (1) a measure is nowhere negative, so \( \nu(A \cap B) \ge 0\), rearranging above $$ \nu(A\cup B) \le \nu(A) + \nu(B) $$

3. Infinite Unions

If \(A_1 \subset A_2 \subset A_3 \subset … \) is an increasing sequence of measurable sets then $$ \nu(\cup_{i=1}^\infty A_i) = \text{lim}_{n\rightarrow\infty}\nu(A_n)$$

Proof

Set \(A_0 := \emptyset \) and \(B_i := A_i \backslash A_{i-1} , i \ge 1\).

Note that \(B_i\) are pairwise disjoint and that $$ A_i = B_1 \cup … \cup B_i \text{ for all } i \ge 1$$ Consequently, $$ \cup_{i=1}^\infty A_i = \cup_{i=1}^\infty B_i $$

Thus
\( \begin{align} \nu( \cup_{i=1}^\infty A_i) &= \sum_{i=1}^\infty \nu(B_i) \\
&= \text{lim}_{i \rightarrow \infty}\sum_{j=1}^\infty \nu(B_j) \\
&= \text{lim}_{n\rightarrow\infty}\nu(A_n) \end{align} \)

4. Infinite Intersections

If \(A_1 \supset A_2 \supset A_3 \supset … \) is an decreasing sequence of measurable sets and if \( \nu(A_i \lt \infty )\) then $$ \nu(\cup_{i=1}^\infty A_i) = \text{lim}_{n\rightarrow\infty}\nu(A_n)$$

Proof

Set \(C_i := A_1 \backslash A_i\).

Note that \(\nu (A_i) + \nu(C_i)=\nu(A_1) \) and that \(\nu(C_i) \le \nu (A_1) \lt \infty \). So then $$ \nu (A_i) = \nu(C_i) - \nu(A_1) $$

Since $$ \cup_{i=1}^\infty C_i = \cup_{i=1}^\infty (A_1 \backslash A_i) = A_1 \backslash (\cap_{i=1}^\infty A_i ) $$

and using the fact that \(\nu(A_1) \lt \infty\) (from definition above)

$$ \nu(\cap_{i=1}^\infty A_i) = \nu(A_1) - \nu(\cup_{i=1}^\infty C_i) $$

Using result from 3. Infinite Unions

\(\begin{align} \nu(\cap_{i=1}^\infty A_i) &= \nu(A_1) - \nu(\text{lim}_{n\rightarrow\infty}C_n) \\
&= \text{lim}_{n\rightarrow\infty}\nu(A_1-C_n) \\
&= \text{lim}_{n\rightarrow\infty}\nu(A_n) \end{align}\)