Relationship between two measures on the same space using dominated convergence

Question

Suppose that $f\in C^\infty_c (\mathbb R, \mathbb R)$ is a real-valued non-negative measurable function on $\mathbb R$ with compact support³:

$$\mathrm{supp}(f) \subseteq [m,M] \,, \tag1$$

and zeroes only in a set of measure zero.

The above function is measurable with respect to two positive measures $d\mu(\lambda)$ and $d\nu(\lambda)$ with $\lambda \in \mathbb R$ such that¹ for some positive non-decreasing function $\varphi$ on $\mathbb R,$

$$ \varphi(m) \le \frac{\int_{\mathbb R} d\mu(\lambda)\ f(\lambda)}{\int_{\mathbb R} d\nu(\lambda)\ f(\lambda)} \le \varphi(M) \,. \tag2$$

From this, I would like to prove² that

$$ d\mu(\lambda) = \varphi(\lambda)\ d\nu(\lambda) \,. \tag3$$

If the above is not true in general, it should at least be true¹ for $\varphi(x) = \exp(\beta x)$ for some $\beta \gt 0.$ I would appreciate a proof for this special case at the very least.

Kindly advise.

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1. Page 36 of these notes on the proof of the Energy-Entropy-Balance (EEB) inequality show that eq. (2) follows from the EEB assumption, $$ \mu(f\log\varphi) \ge \mu(f) \log \frac{\mu(f)}{\nu(f)} \tag4$$ $$ \nu(f\log\varphi) \le \nu(f) \log \frac{\mu(f)}{\nu(f)} \tag5$$ where $\varphi(x) \equiv exp(\beta x)$ and $\mu(f) := \int_{\mathbb R} d\mu(\lambda)\ f(\lambda)$ and $\nu(f):= \int_{\mathbb R} d\nu(\lambda)\ f(\lambda)\,.$

2. It has been suggested that a partition of unity on $f$ and dominated convergence theorem should suffice to prove the above claim. I cannot see how.

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EDIT about notation:

3. I hope it is clear that when I say $\mathrm{supp}(f) \subseteq [m,M] ,$ I mean that $$\inf(\mathrm{supp}(f))\mathrel{\overset{def}{=}} m \qquad \text{and} \qquad \sup(\mathrm{supp}(f)) \mathrel{\overset{def}{=}} M \,. $$

Answer 1

Goal:

We will prove that for the finite measures $\mu$ and $\nu,$ $$ \color{darkblue}{\int_{\mathbb R} \mathrm d\nu(\lambda)\ f(\lambda) = \int_{\mathbb R} \mathrm d\mu(\lambda)\ \frac{1}{\varphi(\lambda)}\ f(\lambda) \tag{∗}}$$ which is equivalent to (3) because $f$ is arbitrary (cf. Radon-Nikodym Derivative).

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Step 1: Bounds on products of compactly supported functions

For a $\varphi$ which is positive and non-decreasing in the interval $[m,M]$ (such as the exponential), we have that $$\forall x \in \mathbb R: \quad \varphi(m) f(x) \le \varphi (x) f(x) \le \varphi(M) f(x) \,. \tag6$$

Choose $\varphi(x) \equiv \exp(\beta x),$ for instance, so that the above reads:

$$ e^{\beta m} f(x) \le e^{\beta x} f(x) \le e^{\beta M} f(x) \,,$$

Or, equivalently, with $w(x):=1/\varphi(x),$ a positive non-increasing function,

$$w(m) f(x) \ge w(x) f(x) \ge w(M) f(x) \,. \tag7$$

Defining the linear functional $\mu(f) := \int_{\mathbb R} \mathrm d\mu(\lambda)\ f(\lambda),$ one rewrites (7) as

$$ w(m) \mu(f) \ge \mu(w f) \ge w(M) \mu(f) \,. \tag8$$

Step 2: (Pointwise) Partition of Unity

Now, consider a sequence of functions $(f_n)_{n \in \mathbb N}$ with compact support $supp(f_n) \subseteq [m_n, M_n]$ and $\sum_n f_n =1$ pointwise (that is, it is a partition of unity) such that for some $\varepsilon\gt 0,$

$$|w(m_n)-w(M_n)| \lt \varepsilon \,. $$

Thus, $f = \sum_n f f_n$ and $supp (ff_n) \subseteq [m_n, M_n] \subset [m,M]$ for sufficiently small $\varepsilon.$ Moreover,

$$ \nu(f) = \nu(\sum_n ff_n) = \sum_n \nu(ff_n)\,. \tag 9$$

The last equality holds because $f\in C^\infty_c (\mathbb R)$ and so the sum is finite.

Similarly, $\mu(wf) = \sum_n \mu(wff_n)$ and hence,

$$ \color{darkblue}{|\nu(f)-\mu(wf)|} \le \sum_n |\nu(ff_n) -\mu(wff_n)| \tag{10}$$

From (8) one immediately sees that,

$$ -\mu(wff_n) \le -w(M_n) \mu(ff_n)\,, \tag{11} $$

and from (the reciprocal of) (2) one sees that

$$ \nu(ff_n) \le w(m_n) \mu(ff_n) \,. \tag{12}$$

Adding (11) and (12), one gets that

$$ \nu(ff_n) - \mu(wff_n) \le \big(w(m_n) - w(M_n)\big) \mu(f_n) $$ $$ \Rightarrow |\nu(ff_n) - \mu(wff_n)| \lt \varepsilon |\mu(ff_n)| \tag{13}$$

Finally, from (10) one gets:

$$ \color{darkblue}{|\nu(f)-\mu(wf)|} \le \sum_n \varepsilon |\nu(ff_n)| \tag{14}$$

Since $\varepsilon$ is arbitrary, one obtains the desired result $ \color{darkblue}{(∗)}.$ Note that we only checked the equality of the (finite!) measures on $f\in C^\infty_c (\mathbb R)$ in which case the sum above has a finite number of finite non-zero terms.

But one can extend this result to the space of all continuous functions which vanish at infinity $C_0 (\mathbb R)$ because the measures are finite and $C^\infty_c (\mathbb R)$ is dense in $C_0 (\mathbb R).$

$$\tag*{$\blacksquare$}$$

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Technical Note:

In order to make sense of the result in $C_0 (\mathbb R),$ we need to make sure that we can sensibly extend our definition of the measures to this set.

It can be shown that the set $\mathfrak B$ of bounded Borel measurable functions is the smallest set $\mathscr F$ such that:

1. $C_0 (\mathbb R) \subseteq \mathscr F,$ and

2. if $(F_n)_{n\in\mathbb N}$ is uniformly bounded (that is $|F_n(x)| \le M\ \forall x,n$ for some number $M \gt 0$), $F_n \in \mathscr F$ and $F_n \to F$ pointwise, then it follows that $F \in \mathscr F.$

This means that the restriction of a finite Borel measure to $C_0 (\mathbb R)$ determines the measure uniquely: In fact, let $F \in \mathscr F$ and $F_n \to F$ as above, then by dominated convergence $$\mu(F) = \lim_{n\to\infty} \mu(F_n) \,.$$

Answer 2

I think this is the result the original poster was hoping for, with, I hope, more precision and a less rambling proof plan.

Theorem Let $\varphi:\mathbb{R}\to\mathbb{R}^+$ be continuous and increasing. Let $\mu$ and $\nu$ be two Radon measures on $\mathbb{R}.$ Suppose that for all non-negative $f\in C_c^\infty(\mathbb{R})$ the following holds:
$$\varphi(m)\, \nu(f) \le \mu(f) \le\nu(f)\, \varphi(M)\, $$ for all $m$, $M$ for which $\mathrm{supp}(f)\subseteq [m,M]$.

Then $\mu$ is absolutely continuous with respect to $\nu$. Futher, the Radon-Nikodym derivative $\frac {d\mu}{d\nu} $ is equal to $\varphi$ except on $\nu$-null sets.

(Here I use the notation $\nu(g)$ to mean $\int_{\mathbb{R}} g d\nu$, and so on. And for sets $A$, $\nu(A)$ to mean $\nu(\chi_A)$, where $\chi_A$ denotes the characteristic function (or indicator function) of $A$.)

Sketch of Proof

Note that for an interval $I=[a,b]$, the hypotheses of the theorem give us $$\mu(I) \le \nu(I) \,\varphi(b),$$ by applying the dominated convergence theorem to test functions $f_n$ converging to the characteristic function of $I$. In greater detail: For each $\epsilon>0$ let $f_n \to \chi_I$ for $f_n\in C_c^\infty(\mathbb{R}),$ with $\mathrm{supp}(f_n)\subset [a-\epsilon,b+\epsilon]$. The hypotheses of the theorem give us $\mu(f_n) \le \nu(f_n) \varphi(b+\epsilon)$. The dominated convergence theorem gives us $\lim_n \mu(f_n) = \mu(\chi_I) = \mu(I)$, and so on. Thus: $\mu(I)\le \nu(I) \varphi(b+\epsilon)$. This holds for all $\epsilon>0$ and $\varphi$ is continuous, so finally $\mu(I)\le \nu(I) \varphi(b)$ as claimed.

First, the argument for absolute continuity: We need to show that $\nu(A)=0$ implies $\mu(A)=0$. By $\sigma$-finiteness it suffices to check this for bounded sets $A \subseteq [-K,K]$. Cover $A$ with the union of finitely many intervals, $B = \bigcup I_i \subseteq [-K,K]$, with $\nu(B) = \epsilon$, for arbitrarily small $\epsilon$.
Then $$ \mu(A) \le \mu(B) \le \nu( B)\, \varphi(K) =\epsilon\, \varphi(K). $$ Since $\epsilon$ is arbitrary, $\mu( A)=0$.

So the Radon-Nikodym derivative of $\mu$ with respect to $\nu$ exists. Call it $\psi$. Our task now is to show that $\psi=\varphi$ everywhere except possibly off of a $\nu$ null set. This seems to me a harder task than establishing $\mu\ll \nu$. Obstacles include: $\nu$ might be singular measure, $\psi$ is only defined $\nu$-a.e., and $\psi$ might not be continuous. Here is one possible approach. There is something called the Lebesgue differentiation theorem, discussed in https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem, and a generalization to arbitrary Borel measures, Lemma 4.1.2 of Ledrappier and Young, ``The metric entropy of diffeomorphisms, Part I...'', Annals of Mathematics 1985, pp 509--539. (See page 524 for the Lemma.) It seems to imply that for $\nu$ almost all $a$, the limit $$\lim_{r\to 0} \frac {\int_{a-r}^{a+r} \psi(x) d\nu} {\nu([a-r,a+r])} = \lim_{r\to0} \frac {\mu([a-r,a+r])} {\nu([a-r,a+r])} = \psi(a).$$ (Notation dictionary: what we call $\psi$ and $\nu$ corresponds to the paper's $g$ and $\mu$.) For each $r>0$ the ratio is bounded below by $\varphi(a-r)$ and above by $\varphi(a+r)$, and $\varphi$ is continuous, so for almost all $a$, we have $\psi(a)=\varphi(a)$.

Comment Like the other proffered solution, this also uses the dominated convergence theorem and (implicitly) partitions of unity.