Guide to Royden and Fitzpatrick’s Real Analysis: The Radon-Nikodym Theorem (Part 1)

Let $(X,\mathcal M, \mu)$ be a $\sigma$-finite measure space; that is, $X$ may be written as a countable union of measurable sets of finite $\mu$-measure. Let $\nu$ be another $\sigma$-finite measure on $(X, \mathcal M)$. Further suppose that $\mu$ and $\nu$ are related to each other by absolutely continuity; in particular, suppose $\mu$ and $\nu$ have the property that

\begin{equation*} \nu(E) = 0 \text{ whenever } \mu(E) = 0 \text{ for all } E \in \mathcal M. \end{equation*}

Should this relationship be true, we say that $\nu$ is absolutely continuous with respect to $\mu$, and we denote this by $\nu \ll \mu$. The Radon-Nikodym Theorem tells us that there exists a nonnegative measurable function $f$ on $X$ such that

\begin{equation} \nu(E) = \int_E f\, d\mu \text{ for all } E\in \mathcal M. \label{eq1}, \end{equation}

and that this function $f$ is unique up to pointwise almost everywhere equivalence.

The Radon-Nikodym theorem is in the same “flavor” as the Riesz Representation Theorem, in the sense that it suggests that certain abstract functions of interest (in this case, measures) may possess a simpler concrete form (in this case, integration of a fixed function over a measurable set).

My aim in this article is to prove this theorem through a deductive line of reasoning which still makes use of the arguments found in Royden and Fitzpatrick’s Real Analysis textbook. I find that by rearranging the arguments in this way, one may find a natural albeit indirect and long path towards our main result which is usually obscured or discarded in favor of the compression and formalization of a proof.

Insights on the recovery of $f$

Our first goal is to construct a candidate function $\hat f$ which we could use as our choice of $f$ in $(\ref{eq1})$. This may be a difficult task using only the abstract properties of measure, so instead, let us try to get insights by tackling a modified form of the problem at hand.

Let us assume that $\nu$ may indeed be written in the form in ( \ref{eq1}) for some nonnegative measurable function $f$, but no other information concerning $f$ is known. Using only the values of $\nu$, is there a way to reliably recover $f$? To reduce the complexity of this problem, let us assume that $\nu(X) = \int_X f\, d\mu < \infty$.

In essence, through $\nu$, we only know what the integral of $f$ is over arbitrary measurable sets. With this in mind, let us forget $\nu$ for now and focus on $f$. Information on the integrals of $f$ is extremely useful, considering that there are many results that allow us to conclude something about a function knowing only its integrals. To wit, here is a simple example which is a consequence of Chebychev’s inequality:

  • If $g$ is a measurable function on $X$, $g$ is nonnegative on $E\in\mathcal M$, and $\int_E g = 0$, then $g = 0$ a.e. on $E$.

This rather simple and intuitive result leads to a very important corollary in the context of our problem.

Let $g_1$ be nonnegative and integrable on $X$. Let $g_2$ be a nonnegative measurable function on $X$. If $\int_E g_1\,d\mu \geq \int_E g_2\,d\mu$ for any $E\in\mathcal M$, then $g_1 \geq g_2$ a.e. on $X$.

Proof. Since $\int_X g_2 \,d\mu \leq \int_X g_1 \,d\mu < \infty$, we conclude $g_1$ and $g_2$ are finite a.e. on $X$ hence $g_1-g_2$ is well-defined and $\int_{E}g_2 \,d\mu < \infty$ for any $E\in\mathcal M$.

By way of contradiction, assume $g_1 < g_2$ on some measurable set $E'$ satisfying $\mu(E') > 0$. As a consequence, $\int_{E'} g_1\,d\mu \leq \int_{E'} g_2\,d\mu$. From the conditions of the proposition, we conclude $\int_{E'} g_1\,d\mu = \int_{E'} g_2\,d\mu$, or $\int_{E'} g_2-g_1\,d\mu = 0$. Hence, $g_2 - g_1 = 0$ a.e. on $E'$; a contradiction.

By the monotonicity of integration, the converse of the above corollary holds. Thus, we have found a characterization for when a nonnegative integrable function on $X$ dominates another nonnegative measurable function a.e. on $X$ using integrals. With this in mind, let us consider the set

\begin{align} \mathcal F &= \left\{ g : X\to [0,\infty] \mid g \text{ measurable }, g\leq f \text{ a.e. on }X \right\} \nonumber \\ &= \left\{ g : X\to [0,\infty]\, \Bigg|\, g \text{ measurable }, \int_E g \leq \nu(E) = \int_E f \text{ for }E\in\mathcal M \right\} \end{align}

We recall that $\int_X f$ is the supremum of all nonnegative simple functions which $f$ dominates (possibly a.e.) on $X$. It is not too difficult to show that $\int_X f$ also coincides with the supremum of the integrals of nonnegative measurable functions over $X$ which $f$ dominates (possibly a.e) on $X$. Therefore,

\begin{equation*} \sup_{g\in\mathcal F} \int_X g = \int_{X} f. \end{equation*}

As a consequence, we can find a sequence $\{g_n\}_{n=1}^\infty$ of functions in $\mathcal F$ such that $\{\int_X g_n\,d\mu\}\to \int_X f \,d\mu$. If, perchance, the sequence converges pointwise to some function $g$ on $X$, then we might be getting closer to our goal.

Indeed, since $f$ dominates each function in $\{g_n\}$ a.e. on $X$, we expect $g\leq f$ or $f-g \geq 0$ a.e. on $X$. Further, if by some coincidence we find that $\int_X g\,d\mu = \lim_{n\to\infty} \int_X g_n \,d\mu = \int_X f\,d\mu$, then $\int_X f-g \,d\mu = 0$. By applying the “simple example” earlier to $f-g$, we conclude $f-g = 0$ a.e. on $X$! Thus $f$ is recovered through $g$. QED.

…But of course, we cannot guarantee such a string of coincidences. So instead, let us construct a sequence that would satisfy our desired properties using $\{g_n\}_{n=1}^\infty$ as our building blocks. For each $n\in\mathbb N$, let us define $\hat f_n = \max\{g_1,g_2,\ldots, g_n\}$. By construction, $\{\hat f_n\}$ is increasing, so by the Monotone Convergence Theorem it converges pointwise to some nonnegative measurable function $\hat f$ on $X$. In addition,

\begin{equation*} \int_X \hat f = \lim_{n\to\infty} \int_X\hat f_n . \end{equation*}

Now, for $n\in\mathbb N$, we know that the functions $g_1, \ldots, g_n$ are each dominated a.e. on $X$ by $f$. As a consequence, we expect $\hat f_n = \max\{g_1, \ldots , g_n \}$ to be dominated a.e. on $X$ by $f$ as well, i.e., $\hat f_n \leq f$ a.e. on $X$. Therefore, the pointwise limit $\hat f$ of $\{f_n\}$ must be dominated a.e. on $X$ by $f$, i.e., $\hat f_n \leq f$ a.e. on $X$. By the monotonocity of integration, $\int_X \hat f \,d\mu \leq \int_X f \,d\mu$.

In addition, since $\hat f_n \geq g_n$ by our construction of $f_n$, we find that

\begin{equation*} \int_X \hat f \,d\mu = \lim_{n\to\infty} \hat f_n \,d\mu \geq \lim_{n\to\infty} g_n = \int_X f\,d\mu. \end{equation*}

We could thus conclude that $\int_X \hat f\,d\mu = \int_X f \,d\mu$. Now, recall that $\hat f \leq f$ a.e. on $X$. By applying our earlier mentioned “simple result” to $f-\hat f$ on $X$, we conclude that $f -\hat f = 0$ a.e. on $X$. That is, $\hat f = f$ a.e. on $X$.

We have now recovered $f$! To summarize, here are the main steps which we employed:

  1. Consider the set \begin{equation*} \mathcal F = \left\{ g: X \to [0,\infty] \,\mid\, g\text{ measurable}, \int_E g\,d\mu \leq \int_E f\,d\mu \text{ for } E\in\mathcal M \right\} \end{equation*} and get a sequence $\{g_n\}$ in $\mathcal F$ such that \begin{equation*} \int_X g_n \,d\mu \to \sup_{g\in\mathcal F}\int_X g\,d\mu= \int_X f\,d\mu. \end{equation*}

  2. From $\{g_n\}$, construct a new sequence $\{\hat f_n\}$ where $\hat f_n = \max\{g_1,\ldots, g_n\}$. Let $\hat f$ be the pointwise limit function of this sequence which is guaranteed to exist by the Monotone Convergence Theorem.

  3. Show that $\hat f \leq f$ a.e. on $X$ and that $\int_X \hat f = \int_X f$, thereby conclude that $\hat f = f$ a.e. on $X$.

We have thus found a procedure to get $f$ using only its integrals. Using this as our guide, in the next article, we shall replace the integration of $f$ with the evaluation of an arbitrary measure, and see which among the steps above could be retained or modified in order to prove the Radon-Nikodym theorem.

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