Measure theory 101

Measures and spaces

Definition 1(Definition).

Consider a set $X$ and a $\sigma$-algebra $\mathcal{F}$ on $X$. The $(X,\mathcal{F})$ is called a measurable space. The elements of $\mathcal{F}$ are called measurable sets within the space.

Definition 2(Definition).

Let $(X,\mathcal{F})$ be a measure space. A set-theoretic function $\mu :\mathcal{F}\to \mathbb{R}$ is called a measure if the following hold:

1. $\mu (E)\ge 0$ for all $E\in \mathcal{F}$.
2. $\mu (\varnothing )=0$.
3. For all countable collections $\{E_k{\}}_{k=1}^{\infty }$ of pairwise disjoint sets in $\mathcal{F}$,

$$\mu \left(\bigcup _{k=1}^{\infty }{E}_k\right)=\sum _{k=1}^{\infty }\mu (E_k).$$

Definition 3(Definition).

A measure space is a measurable space $(X,\mathcal{F})$ with a measure $\mu$.

Definition 4(Definition).

Let $(X,\Sigma )$ and $(Y,\Delta )$ be measurable spaces. A function $f:X\to Y$ is said to be measurable if for every $E\in \Delta$ the pre-image of $E$ under $f$ is in $\Sigma$. In other words, $f^{-1}(E)\in \Sigma$ for all $E\in \Delta$.

This is in direct analogy the the definition of a continuous function between topological spaces, where the preimage of any open set is open.

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Lebesgue Integration

The Dirichlet function exhibits one of the principal drawbacks of the Riemann integral: a uniformly bounded ¹, increasing sequence of Riemann integrable functions on a closed, bounded interval may converge pointwise to a function that is not Riemann integrable. The Lebesgue integral does not have this shortcoming.

In the following text, the domains of functions is assumed to be a measure space with measure $m$.

Some definitions:

1. A function is said to be finitely supported if it vanishes on the complement of a set of finite measure.
2. A function is said to be simple if it takes on only finitely many values. If ${\lambda }_1,\dots ,{\lambda }_n$ are the values taken by a simple function $\phi$, $\phi (x)={\sum }_{i=1}^n{\lambda }_i{\chi }_{E_i}(x)$, where $E_i\equiv {\phi }^{-1}({\lambda }_i)$ and ${\chi }_{E_i}$ is the indicator function of $E_i$.

Definition 5(Definition).

Let $\psi :E\to \mathbb{R}$ be a finitely supported, simple function. Then the integral of $\psi$ over $E$, ${\int }_E\psi dm$, is defined as follows: if $\psi$ is identically zero, define ${\int }_E\psi dm\equiv 0$. Otherwise, let ${\lambda }_1,\dots ,{\lambda }_n$ be the finite number of nonzero values taken by $\psi$ and define

$${\int }_E\psi dm=\sum _{i=1}^n{\lambda }_i\cdot m(E_i),$$

where $E_i={\psi }^{-1}({\lambda }_i)$.

Definition 6(Definition).

A bounded measurable function $f:E\to \mathbb{R}$ defined on a measurable set $E$ is said to be Lebesgue integrable if

$$\underset{\phi \le f}{\sup }{\int }_E\phi dm=\underset{f\le \psi }{\inf }{\int }_E\psi dm$$

where the infimum and supremum are taken over finitely supported, simple functions on $E$. The common value of the infimum and supremum is called the Lebesgue integral of $f$ over $E$, denoted by ${\int }_{E}fdm$.

Analogous to the definition of the Riemann integral.

If $f:[a,b]\to \mathbb{R}$ is a bounded, Riemann integrable function, then it is Lebesgue integrable and the two integrals are equal.

If $f:E\to \mathbb{R}$ is a bounded, finitely supported, measurable function, then it is Lebesgue integrable.

Theorem 7(Bounded convergence theorem).

Assume $m(E)<\infty$ and $\{f_n:E\to \mathbb{R}\}$ is a uniformly pointwise bounded sequence of measurable functions (note that these conditions make the $f_n$ Lebesgue integrable). If $\{f_n\}\to f$ pointwise on $E$, then

$$\underset{n\to \infty }{\lim }{\int }_E{f}_{n}dm={\int }_{E}fdm.$$

Definition 8(Definition).

The integral of a nonnegative measurable function $f:E\to [0,\infty ]$ is defined by

$${\int }_{E}fdm=\underset{0\le h\le f}{\sup }{\int }_{E}hdm,$$

where the supremum is taken over bounded, finitely supported, measurable functions $h$.

Note the the supremum may be infinity, and is always defined.

Theorem 9(Fatou's Lemma).

If $\{f_n:E\to [0,\infty ]\}$ is a sequence of nonnegative, measurable functions and $\{f_n\}\to f$ pointwise on $E$, then

$${\int }_{E}fdm\le \underset{n\to \infty }{\mathrm{liminf}}{\int }_E{f}_{n}dm.$$

Definition 10(Definition).

A nonnegative, measurable function $f:E\to [0,\infty ]$ is said to be integrable provided that

$${\int }_{E}fdm<\infty .$$

Definition 11(Definition).

A measurable function $f:E\to \mathbb{R}$ is said to be integrable provided that ${\int }_E|f|dm<\infty$, and for such a function, the integral of $f$ over $E$ is defined by

$${\int }_{E}fdm={\int }_E{f}^{+}dm-{\int }_E{f}^{-}dm.$$

Theorem 12(The integral comparison test).

If $f:E\to \mathbb{R}$ is a measurable function, and there is a nonnegative, integrable function $g:E\to \mathbb{R}$ that dominates $f$ on $E$, in the sense that $|f|\le g$ on $E$, then $f:E\to \mathbb{R}$ is integrable and

$$\left|{\int }_{E}fdm\right|\le {\int }_E|f|dm.$$

Dominated convergence theorem

Theorem 13(Dominated convergence theorem).

Let $\{f_n:E\to \mathbb{R}\}$ be a sequence of measurable functions. Assume that there is an integrable function $g:E\to [0,\infty ]$ that dominates $\{f_n\}$ on $E$, in the sense that $|f_n|\le g$ on $E$ for all $n$. If $\{f_n\}\to f$ pointwise on $E$, then $f:E\to \mathbb{R}$ is integrable and

$$\underset{n\to \infty }{\lim }{\int }_E{f}_{n}dm={\int }_{E}fdm$$

The following theorem replaces $m$ with the standard Lebesgue measure. It can be proven using the dominated convergence theorem.

Theorem 14(Differentiating under the integral sign).

Assume

1. $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$,
2. the map $x\mapsto f(x,t)$ be measurable for all $t\in \mathbb{R}$,
3. $f(x,t_0)$ is Lebesgue integrable for some $t_0$,
4. $\frac{\partial }{\partial t}f(x,t)$ exists for all $(x,t)\in {\mathbb{R}}^2$, and
5. there exists Lebesgue integrable $g:\mathbb{R}\to \mathbb{R}$ such that ${\left|\frac{\partial f}{\partial t}\right|}_{(x,t)}\le g(x)$ for all $(x,t)\in {\mathbb{R}}^2$.

Then, the map $x\mapsto f(x,t)$ is integrable for all $t$, and the function $F$ defined by

$$F(t)\equiv {\int }_{\mathbb{R}}f(x,t)dx$$

is differentiable with derivative

$$F^{\prime }(t)=\frac{d}{dt}{\int }_{\mathbb{R}}f(x,t)dx={\int }_{\mathbb{R}}\frac{\partial }{\partial t}f(x,t)dx.$$

Footnotes

1. A sequence of real-valued functions is said to be uniformly bounded if there exists $M$ such that $|f_n(x)|\le M$ for all $x\in E$ and $n\in \mathbb{N}$. ↩