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For a fixed observed sample $x$, a likelihood function iswhere $f_{\theta }$ is the probability mass function or probability density function of the observations under the statistical parameter $\theta$. All densities must be defined relative to the same parameter-independent dominating measure. The sample is fixed; the parameter is the function's argument. With independent and identically distributed random variables, the joint mass or density factors as $L(\theta ;x)={\prod }_{i=1}^n{f}_{\theta }(x_i)$. This definition and the distinction between sampling probability and parameter uncertainty are explained in MIT 18.05, Reading 10b, sections 3 and 3.2.

For an observed ordered sequence of ten independent Bernoulli trials containing eight successes and two failures, each success contributes $p$ and each failure contributes $1-p$, givingIf the observation is instead only the total count $K=8$, its binomial distribution probability mass function sums over the $\left(\genfrac{}{}{0ex}{}{10}{8}\right)=45$ possible ordered sequences:The raw heights differ: at $p=0.8$ they are approximately $0.0067108864$ and $0.301989888$. Their maximum-likelihood estimates are identical. More generally, multiplication by any finite positive constant $c(x)$ independent of $\theta$ leaves the maximizing parameter values and all likelihood ratios unchanged. Multiplication by a parameter-dependent factor can change the MLE; dropping such a factor is not justified. Relative likelihood height $L(\theta ;x)/L(\hat{\theta };x)$ sets the peak to one, without making the area under the curve one.

Normalization of $f_{\theta }$ is over possible data values for each fixed parameter. It imposes no requirement that ${\int }_{\Theta }L(\theta ;x)d\theta =1$. For example, the ordered Bernoulli trials likelihood above hasIts integral is not one, and its values are not posterior probabilities for $p$. When the Bayesian posterior is defined, Bayes' theorem multiplies the likelihood by a prior density and normalizes that product. Choosing to normalize a likelihood function alone is an additional choice, such as using a uniform prior in those parameter coordinates, rather than part of the definition of likelihood.

For observations with a probability density function, the probability of an exact sample point is zero. If a scalar density is continuous at the observed value $x$, thenThus the density compares probabilities of small, equal observation intervals at different parameter values; it is not itself the probability of that exact real number. The equal-width factor is independent of the parameter. This gives visual intuition for density likelihoods, as in MIT 18.05, Reading 10b, section 4. The normal distribution example in maximum-likelihood estimator makes the comparison explicit. If only rounded intervals are recorded, their integrated probabilities give the exact observation likelihood; using midpoint densities is an approximation.

We construct the density and singular part, rather than assuming the desired Lebesgue decomposition theorem. Put $\rho =\mu +\nu$, a finite dominating measure, and use the real Hilbert space $L^2(\rho )$. The linear functionalis well defined on its equivalence classes, since a $\rho$-null set is $\nu$-null. The Cauchy-Schwarz inequality and $\nu \le \rho$ giveBy the Riesz representation theorem, there is $h\in L^2(\rho )$ with $T(g)=\int ghd\rho$. Indicators of all measurable sets belong to this Hilbert space, soThe nonnegativity of these two measures forces $0\le h\le 1$ almost everywhere. For instance, if $\rho (\{h\le -1/n\})>0$, its $\nu$-measure would be negative; the analogous sets $\{h\ge 1+1/n\}$ contradict positivity of $\mu$. Their countable union excludes every violation. Modify $h$ on its measurable $\rho$-null exceptional set so that these inequalities hold everywhere. This is the Hilbert-space construction of dominated measure densities.

DefineThen $B\in \Sigma$, $f$ is a nonnegative measurable function, and $\mu (B)={\int }_B(1-h)d\rho =0$. The identity for integration against $\mu$ follows first for indicators, then for nonnegative simple functions and finally for all nonnegative functions by the monotone convergence theorem. Applying it to $f{1}_A$ givesIn particular ${\int }_{\Omega }fd\mu =\nu (\Omega \setminus B)<\infty$, so $f\in L^1(\mu )$. Additivity on the disjoint measurable sets $A\setminus B$ and $A\cap B$ now givesThis is the Lebesgue decomposition from a sum-measure density: the first measure has a density, and the second is concentrated on a $\mu$-null set, so is mutually singular with $\mu$.

The absolute continuity of measures condition $\nu \ll \mu$ means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If this holds, then $\nu (B)=0$, eliminating the singular part above. Conversely, a measure of the form $\nu (A)={\int }_{A}fd\mu$ vanishes on every $\mu$-null set. ThereforeThus absolute continuity is exactly the absence of the singular part in this construction.

Construct the densities using the Riesz representation theorem for a Hilbert space, rather than assuming the measure-decomposition conclusion. Put $\rho =\mu +\nu$, a finite dominating measure, and work in the real L2 space $L^2(\rho )$. The linear functionalis well-defined on its equivalence classes because $\nu \le \rho$. By the Cauchy-Schwarz inequality,Thus the Riesz representation theorem supplies $h\in L^2(\rho )$ such that $L(g)=\int ghd\rho$. Since $\rho$ is finite, $h$ is also integrable. Testing $g=1_A$ givesThe first identity and nonnegativity of $\nu$ rule out a positive-measure set where $h<0$; for example, test $\{h\le -1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge 1+1/n\}$. Modify a measurable representative on a $\rho$-null set so that $0\le h\le 1$ everywhere. This is the Hilbert-space construction of dominated measure densities.

SetThen $B\in \Sigma$ and $\mu (B)=0$. The indicator identities for $\mu$ extend to integrals of nonnegative measurable functions by approximation with simple functions and the monotone convergence theorem. Consequently,and, for every $A$,HenceThis proves Lebesgue decomposition from a sum-measure density. The argument includes the cases of zero measures; there is no division on $B$, where the denominator vanishes.

Absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If $\nu \ll \mu$, then $\nu (B)=0$ in the construction above, so the singular term vanishes. Conversely, if $\nu (A)={\int }_{A}fd\mu$ with a nonnegative integrable $f$, that integral is zero on every $\mu$-null set. ThusThis is the finite-measure Radon-Nikodym theorem as a consequence of the proved construction.

Finally restrict the base measure to the sub-sigma-algebra ${\Sigma }_0$. For a real $f\in L^1(\mu )$, write $f=f^{+}-f^{-}$, with both parts nonnegative and integrable. On ${\Sigma }_0$ define finite positive measuresThey are absolutely continuous with respect to $\mu {|}_{{\Sigma }_0}$. The result just proved gives nonnegative, ${\Sigma }_0$-measurable densities $g_{\pm }$ satisfyingSet $f_0=g_{+}-g_{-}$. Since $\int (g_{+}+g_{-})d\mu =\int |f|d\mu <\infty$, this is in $L^1(\Omega ,{\Sigma }_0,\mu )$, and subtraction givesFor complex $f$, apply the real construction separately to its real and imaginary parts and combine the two results. This is conditional expectation from finite-measure densities. The resulting conditional expectation is unique almost everywhere: for two real versions, their difference has integral zero on its positive and negative level sets in ${\Sigma }_0$, forcing it to vanish; apply this to both components in the complex case. No probability normalization of $\mu$ is required.

Let $\Delta =\{(x,y):x=y\}$. For any transport plan $\pi$ and any Borel set $A$,Both rectangles are contained in ${\Delta }^c$, soThus $\inf \mathbb{K}\ge {\sup }_A|\mu (A)-\nu (A)|$.

To attain this bound, use the dominating measure $\rho =\mu +\nu$ and Radon-Nikodym derivatives $f=d\mu /d\rho$, $g=d\nu /d\rho$. Define the common measure and residual mass bySince $\int (f-g)d\rho =0$, the positive and negative parts have equal mass. Taking $A=\{f>g\}$ givesIf $r=0$, then $\mu =\nu$ and the diagonal transport plan has zero cost. If $r>0$, put $\beta =\mu -\alpha$, $\gamma =\nu -\alpha$ and defineThe residual measures each have mass $r$, so the marginal distributions of ${\pi }^{†}$ are $\alpha +\beta =\mu$ and $\alpha +\gamma =\nu$, and its total mass is $1-r+r=1$. They are mutually singular measures, concentrated respectively on $\{f>g\}$ and $\{g>f\}$. Their product measure therefore gives no mass to $\Delta$, and $\mathbb{K}({\pi }^{†})=r$. This is a maximal coupling.

In the paper's convention for the zero-mass signed measure $\mu -\nu$, the answer isThe repository's total variation distance uses the supremum itself. The paper's formula $2{\sup }_A|\sigma (A)|$ agrees with the usual total variation norm when $\sigma$ has total mass zero, as here; it is not the usual norm for an arbitrary positive measure. All suprema above are over Borel sets.

The total variation distance isIf both measures have densities $p,q$ with respect to a common dominating measure, then $d_{\mathrm{TV}}(\mu ,\nu )=\frac{1}{2}\int |p-q|$.