Let
be
a
-finite measure
space; that is,
may be written as a
countable union of measurable sets of finite
-measure. Let
be another
-finite measure
on
.
Further suppose that
and
are related to each
other by **absolutely continuity**; in particular, suppose
and
have the property
that

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

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

The Radon-Nikodym theorem is in the same â€śflavorâ€ť 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 Fitzpatrickâ€™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

Our first goal is to construct a candidate function which we could use as our choice of in . 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
may indeed be
written in the form in
(
) for some
nonnegative measurable function
, but no other
information concerning
is known. *Using
only the values of*
, *is there a way
to reliably recover*
? To reduce the
complexity of this problem, let us assume that
.

In essence, through , we only know what the integral of is over arbitrary measurable sets. With this in mind, let us forget for now and focus on . Information on the integrals of 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 Chebychevâ€™s inequality:

- If is a measurable function on , is nonnegative on , and , then a.e. on .

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

Let be nonnegative and integrable on . Let be a nonnegative measurable function on . If for any , then a.e. on .

*Proof.* Since
,
we conclude
and
are finite a.e. on
hence
is well-defined
and
for any
.

By way of contradiction, assume on some measurable set satisfying . As a consequence, . From the conditions of the proposition, we conclude , or . Hence, a.e. on ; 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 dominates another nonnegative measurable function a.e. on using integrals. With this in mind, let us consider the set

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

As a consequence, we can find a sequence of functions in such that . If, perchance, the sequence converges pointwise to some function on , then we might be getting closer to our goal.

Indeed, since dominates each function in a.e. on , we expect or a.e. on . Further, if by some coincidence we find that , then . By applying the â€śsimple exampleâ€ť earlier to , we conclude a.e. on ! Thus is recovered through . QED.

â€¦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 as our building blocks. For each , let us define . By construction, is increasing, so by the Monotone Convergence Theorem it converges pointwise to some nonnegative measurable function on . In addition,

Now, for , we know that the functions are each dominated a.e. on by . As a consequence, we expect to be dominated a.e. on by as well, i.e., a.e. on . Therefore, the pointwise limit of must be dominated a.e. on by , i.e., a.e. on . By the monotonocity of integration, .

In addition, since by our construction of , we find that

We could thus conclude that . Now, recall that a.e. on . By applying our earlier mentioned â€śsimple resultâ€ť to on , we conclude that a.e. on . That is, a.e. on .

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

Consider the set and get a sequence in such that

From , construct a new sequence where . Let be the pointwise limit function of this sequence which is guaranteed to exist by the Monotone Convergence Theorem.

Show that a.e. on and that , thereby conclude that a.e. on .

We have thus found a procedure to get using only its integrals. Using this as our guide, in the next article, we shall replace the integration of 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.