Measures §

Countable additivity implies countable subadditivity: for any (not necessarily disjoint) $A_i \in \mathcal{F}$, $\mu\!\left(\bigcup_i A_i\right) \leq \sum_i \mu(A_i)$

It also implies monotonicity: if $A \subseteq B$ with $A, B \in \mathcal{F}$, then $\mu(A) \leq \mu(B)$.

A measure is finite if $\mu(X) < \infty$, $\sigma$-finite if $X = \bigcup_n A_n$ with $\mu(A_n) < \infty$, and a probability measure if $\mu(X) = 1$.

The Lebesgue measure is the canonical example: it extends the intuitive notion of length to all Lebesgue-measurable subsets of $\mathbb{R}$. It is $\sigma$-finite because $\mathbb{R} = \bigcup_n (-n, n)$ and each interval has finite measure.

The counting measure assigns to each set its cardinality. It is complete (every subset of a null set is null, since the only null set is $\varnothing$), but it is $\sigma$-finite only on countable sets.

The Dirac measure $\delta_a$ concentrates all mass at a single point $a$. It is the measure-theoretic analogue of a point mass in physics.

A probability measure is any measure with total mass 1. The Gaussian (normal) measure is a probability measure with density $\frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)}$ with respect to Lebesgue measure.

Lebesgue outer measure on $\mathbb{R}$: $\lambda^*(A) = \inf \left\{ \sum_{i=1}^{\infty} (b_i - a_i) : A \subseteq \bigcup_{i=1}^{\infty} (a_i, b_i) \right\}$

This is the “infimum of total lengths of countable interval covers.” It is defined for every subset of $\mathbb{R}$, but is only countably additive on the Lebesgue measurable sets. The Carathéodory construction then extracts the largest sigma-algebra (the Lebesgue σ-algebra) on which $\lambda^*$ is a genuine measure.

Intuition. The Radon–Nikodym theorem says that if one measure is “absolutely continuous” with respect to another, it is obtained by weighting the base measure with a density function. In probability, if $P$ and $Q$ are two probability measures with $Q \ll P$, then $\frac{dQ}{dP}$ is the likelihood ratio.

Example. On $(\mathbb{R}, \mathcal{L}, \lambda)$, the measure $\nu(A) = \int_A e^{-x^2} \, d\lambda$ has Radon–Nikodym derivative $\frac{d\nu}{d\lambda} = e^{-x^2}$.

Mutual singularity and the Lebesgue decomposition. The complementary relation to absolute continuity is mutual singularity: $\nu \perp \mu$ means there exists $A \in \mathcal{F}$ with $\nu(A) = 0$ and $\mu(X \setminus A) = 0$ — the two measures “live on disjoint parts” of $X$. The Lebesgue decomposition theorem states that for any $\sigma$-finite $\nu$ on a $\sigma$-finite $(X, \mathcal{F}, \mu)$, there is a unique decomposition $\nu = \nu_{\mathrm{ac}} + \nu_{\mathrm{s}}$ where $\nu_{\mathrm{ac}} \ll \mu$ (the absolutely continuous part, given by a density $\frac{d\nu_{\mathrm{ac}}}{d\mu}$) and $\nu_{\mathrm{s}} \perp \mu$ (the singular part). In probability, this separates a distribution into its continuous (density) and discrete (point-mass) components.

The theorem fails without $\sigma$-finiteness: the counting measure on an uncountable set is not $\sigma$-finite, and there exist measures absolutely continuous with respect to it that have no density.

Tonelli’s theorem (see Lebesgue integral) states that for nonnegative measurable $f$ on $X \times Y$: $\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left(\int_Y f(x,y) \, d\nu(y)\right) d\mu(x)$

Fubini’s theorem strengthens this: if $f$ is integrable (i.e., $\int_{X \times Y} |f| \, d(\mu \times \nu) < \infty$), then the iterated integrals exist, are finite, and equal the double integral. Tonelli applies to all nonnegative functions (the integral may be $\infty$); Fubini requires the integrability hypothesis.

Together they justify iterated integration and form the foundation of multivariable integration in the Lebesgue setting.

Example. Lebesgue measure on $\mathbb{R}^2$ is $\lambda \times \lambda$, the product of 1-dimensional Lebesgue measure with itself. The area of a rectangle $(a,b) \times (c,d)$ is $(b-a)(d-c)$.

Step 1: Identify the density.

By the Radon–Nikodym theorem, since $\nu \ll \lambda$ (the integrand is nonnegative and measurable), the Radon–Nikodym derivative is the integrand itself: $\frac{d\nu}{d\lambda}(x) = 1 + x^2$

Step 2: Compute $\nu([0,1])$.

$\nu([0,1]) = \int_0^1 (1 + x^2) \, dx = \left[x + \frac{x^3}{3}\right]_0^1 = 1 + \frac{1}{3} = \frac{4}{3}$