2025-09-10
\[ \def\R{\mathbb{R}} \def\B{\mathscr{B}} \]
Simplifies some arguments
Unifies treatment
Resolves some difficulties with infinity
Given \(\mathcal{C} \subseteq 2^\Omega\), the \(\sigma\)-field generated by \(\mathcal{C}\) is the intersection of all \(\sigma\)-fields containing \(\mathcal{C}\)
Exercise
If \(\Omega = \{a, b, c\}\) and \(\mathcal{C} = \{ \{a, b\} \}\), what is \(\sigma(\mathcal{C})\)?
Theorem
If \(\mathcal{C}\) is a ring of sets, i.e.
and \(\mu: \mathcal{C} \to \R\) is a pre-measure, i.e.
then there exists a measure \(\mu': \sigma(\mathcal{C}) \to \R\).
Exercise
Let \(\Omega\) be countable with any \(\mathscr{F}\), define \(\mu(A)\) as the number of elements of \(A\). Show \(\mu\) is \(\sigma\) finite.
Theorem
There exists a unique \(\sigma\)-finite measure \(\mu\) on \((\mathbf{R},\mathscr{B}(\mathbf{R}))\) such that for any \(a\leq b\) with \(a,b\in \mathbf{R}\), \[ \mu ((a,b])=b-a \]
Lemma
If \(f: \R^n \to \R^k\) is continuous, then \(f\) is \(\B(\R^n)\) measurable.
Corollary
If \(f: \Omega \to \R\) is measurable and \(g: \R \to \R\) is continuous, then \(g \circ f\) is measurable.
Corollary
If \(f: \Omega \to \R\) and \(g: \Omega \to \R\) are measurable, then \(f+g\) is measurable.
Exercise
Show that if \(f\) is simple, then \(f\) is measurable.
Exercise
Show that if \(f\) and \(g\) are simple, then
If \(f \geq 0\) a.e., then \(\int f d\mu \geq 0\)
\(\forall a \in \R\) , \(\int a f d\mu = a \int f d\mu\)
\(\int (f + g) d\mu = \int f d\mu + \int g d \mu\)
Exercise
Show the second equality above.
Exercise
Show that for all bounded \(f\) and \(g\) that vanish outside a finite measure set,
If \(f \geq 0\) a.e., then \(\int f d\mu \geq 0\)
\(\forall a \in \R\) , \(\int a f d\mu = a \int f d\mu\)
\(\int (f + g) d\mu = \int f d\mu + \int g d \mu\)
Exercise
Show that for all \(f \geq 0\),
If \(f \geq 0\) a.e., then \(\int f d\mu \geq 0\)
\(\forall a \in \R\) , \(\int a f d\mu = a \int f d\mu\)
\(\int (f + g) d\mu = \int f d\mu + \int g d \mu\)
\[ \int f d\mu = \int f^{+} d\mu - \int f^{-} d\mu \]
Exercise
Show that for all measurable \(f\)
If \(f \geq 0\) a.e., then \(\int f d\mu \geq 0\)
\(\forall a \in \R\) , \(\int a f d\mu = a \int f d\mu\)
\(\int (f + g) d\mu = \int f d\mu + \int g d \mu\)
Lemma
Suppose that \(\{E_{n}\}\) is a monotone sequence of sets. Then \[ \mu \left( \lim_{n\rightarrow \infty}E_{n}\right) =\lim_{n\rightarrow \infty }\mu (E_{n}). \]
Lemma
If \(f_n:\Omega \to \mathbf{R}\) are measurable, \(f_{n}\geq 0\), and for each \(\omega \in \Omega\), \(f_{n}(\omega )\uparrow f(\omega )\), then \(\int f_{n}d\mu \uparrow \int fd\mu\) as \(n\rightarrow \infty\)
Exercise
Suppose \(g_n: \Omega \to \R\) are measurable and \(g_n \geq 0\). Show that \[ \int \sum_{n=1}^\infty g_n d\mu = \sum_{n=1}^\infty \int g_n d\mu \]
Lemma
If \(f_n:\Omega \to \R\) are measurable, \(f_{n}\geq 0\), then \[ \int \left( \text{liminf}_{n\rightarrow \infty }f_{n}d\mu \right) \leq \text{liminf}_{n\rightarrow \infty }\int f_{n}d\mu \]
Lemma
If \(f_n:\Omega \to \R\) are measurable, and for each \(\omega \in \Omega\), \(f_{n}(\omega )\rightarrow f(\omega ).\) Furthermore, for some \(g\geq 0\) such that \(\int gd\mu <\infty\), \(|f_{n}|\leq g\) for each \(n\geq 1\). Then, \(\int f_{n}d\mu \rightarrow \int fd\mu\)
Exercise
Show that if \(g_n: \Omega \to \R\) are measurable and \(\sum_{n=1}^\infty \int |g_n| d\mu < \infty\), then \(\sum_{n=1}^\infty |g_n(x)| < \infty\) almost everywhere and if \(g(x) = \sum_{n=1}^\infty g_n(x)\), then \[ \lim_{N \to \infty} \int \left\vert g(x) - \sum_{n=1}^N g_n(x) \right\vert d\mu(x) = 0 \]
Theorem
Let \((\Omega,\mathscr{F},\mu)\) be a measure space, and let \(\nu\) and \(\mu\) be \(\sigma\)-finite measures defined on \(\mathscr{F}\) and \(\nu \ll \mu\). Then there is a nonnegative measurable function \(f\) such that for each set \(A\in \mathscr{F}\), \[ \nu (A)=\int_{A}fd\mu \] For any such \(f\) and \(g\), \(\mu (\{\omega \in \Omega:f(\omega )\neq g(\omega )\})=0\)