A set \(X\) with a distance function \(d: X \times X \rightarrow \mathbb{R}^+\) that satisfies the following:
A topological space is a pair \((X, T)\) where \(T \subset \mathcal{P}(X)\) (called a topology on \(X\)) that satisfies the following:
Sets in \(T\) are called open. Compliments of sets in \(T\) are called closed.
If \(d\) is a metric on \(X\) then sets of the form \(B_\epsilon(x) = \{y \in X | d(x,y) < \epsilon\}\) generate a topology in which \(B_\epsilon(x)\) is the ball of radius \(\epsilon\) around \(x.\)
Let \(X\) be a set with a \(\sigma\)-algebra \(\mathcal{A}\). That is, \(\mathcal{A}\) is a subset of \(\mathcal{P}(X)\) that satisfies the following:
A measure on \((X,\mathcal{A})\) is a function \(\mu: \mathcal{A} \rightarrow \mathbb{R}^+\) such that
The triple \((X, \mathcal{A}, \mu)\) is a measure space.
The Borel measure is the measure on the \(\sigma\)-algebra generated by open sets on \(\mathbb{R}\) such that \(\mu((a,b)) = b - a\) for any open interval \((a,b) \subset \mathbb{R}\).
The Lebesgue measure is the completion of the Borel measure, meaning if any set \(A\) has measure \(0\) then every subset of \(A\) is measurable and also has measure \(0.\)
The classic example is to create equivalence classes \(a \equiv b\) if \(a - b \in \mathbb{Q}\). Choose one representative from every class in \([0,1)\) and create a set \(A\). For all rationals in \(q \in [0,1)\) we can create a disjoint class \(A_q = \{z : z = a + q\) or \(z = a + q - 1, 0 \leq z < 1,\) for all \(a \in A\}.\)
Every class \(A_q\) must have the same measure. They are a countable collection of disjoing sets, so, \(1 = \mu([0,1)) = \mu(\bigcup_{q \in \mathbb{Q}}A_q) = \sum_{q \in \mathbb{Q}}\mu(A_q).\)
If \(\mu(A) = 0\) we get \(1 = 0\) but if \(\mu(A) > 0\) we get \(1 = \infty,\) so \(A\) cannot be measurable.
Let \((X,T)\) and \((Y,U)\) be topolotical spacted. Then function \(f:X \rightarrow Y\) is continuous if \(f^{-1}(A) \in T\) for every \(A \in Y.\)
Let \((X,\mathbb{A},\mu)\) be a measure space and \(Y\) have \(\sigma\)-algebra \(\mathbb{B}.\) The function \(f: X \rightarrow Y\) is measurable if for every \(B \in \mathbb{B},\) \(f^{-1}(B) \in \mathbb{A}.\)
An \(L^+\) simple function is a function of the form \(\sum_{i=1}^\infty a_i 1_{A_i}\) where \(a_i \in \mathbb{R}^+\) and \(A_i \in \mathbb{A}\).
The Lebesgue Intebral of a function \(f \in L^+\) is \(\int f d\mu = \sum_{i=1}^\infty a_i \mu(A_i)\)
For a measurable function \(f: X \rightarrow \mathbb{R}^+,\) \(\int f d\mu = sup\left\{ \int \phi d\mu : \phi \leq f, \phi \in L^+ \right\}.\)
For any function \(f,\) break up \(f\) into \(f^+\) where \(f\) maps to positive values and \(f^-\) where \(f\) maps to negative values. Then \(\int f d\mu = \int f^+ d\mu - \int |f^-| d\mu.\)
A probably measures is a measure \(P\) on a set \(X\) such that \(P(X) = 1.\) If \(f\) is a measurable function \(f: X \rightarrow \mathbb{R},\) then \(\int f dP = E[f].\)