To what extent are measure, topology, distances / norms related?

Question

I m learning about measure theory and I m struggling to make bounds between those different topics.

In fact, in topology we saw that a distance induces a topology. But it appears that a measure is thinner than a topology, meaning that the set of measurables is included in the open set, despite that I m not imagining a measure without a distance!

Can you give some insights on this topic?

Answer 1

For a topological space $X$, one often considers the $\sigma$-algebra generated by the open sets of $X$ (sometimes called the $\sigma$-algebra of Borel sets in $X$ or the Borel $\sigma$-algebra). Of course, this $\sigma$-algebra contains every open set in $X$ by definition. This is a pretty standard construction for topological spaces, generalizing the Borel sets of $\mathbb R$.

A $\sigma$-algebra on $X$ will always contain a topology: the trivial (concrete) topology $\lbrace \emptyset ,X\rbrace$. But if $X$ is a topological space, we can certainly define $\sigma$-algebras on $X$ that have nothing to do with the open sets in $X$. Also note that

• We cannot expect a $\sigma$-algebra to satisfy the axioms of a topology (since arbitrary unions do not have to lie in the $\sigma$-algebra).
• If a $\sigma$-algebra does satisfy the axioms of a topology, then the topology it induces will be rather weird. Specifically, for a $\sigma$-algebra $\mathcal A$ we require $A\in\mathcal A\Rightarrow A^c\in\mathcal A$. If $\mathcal A$ is a topology, this implies that every closed set is open (and vice versa).

In conclusion, it does not make sense to say that $\sigma$-algebras (of measurable sets) are thinner than topologies. This is somehow partially true, insofar as $\sigma$-algebras do not generally contain arbitrary unions. But on the other hand, $\sigma$-algebras contain complements -- which cannot be said of topologies. In fact, when there is a natural relationship between topologies and $\sigma$-algebras -- that is, when we define the Borel $\sigma$-algebra on a topological space to be generated by the topology -- the $\sigma$-algebra will be thicker, so to speak, than the topology (by construction).

Lesbegue measurable sets. You are also interested in the Lesbegue measurable sets on $\mathbb R$. This is a larger $\sigma$-algebra than the Borel sets on $\mathbb R$. What do the Lesbegue measurable sets have to do with the topology on $\mathbb R$? Is there some sort of generalization of the Lesbegue measurable sets to other topological contexts (as with Borel sets)?

These questions are complicated since the Lesbegue $\sigma$-algebra is complicated. But again, since the Lesbegue $\sigma$-algebra on $\mathbb R$ contains the Borel $\sigma$-algebra in $\mathbb R$, it contains in particular all open sets in $\mathbb R$. Here the $\sigma$-algebra is, once again, thicker than the topology (in the sense of larger). As for the second question, Rudin (Real and Complex Analysis, chap. 2) uses the Riesz Representation Theorem (very difficult, but highly beautiful, theorem!) to give a construction of Lesbegue-like $\sigma$-algebras on locally compact Hausdorff spaces. The Lesbegue $\sigma$-algebra is a special case.

Topologies induced by $\sigma$-algebras. In the comment to your own question, you ask the great question of whether every topology is induced by a $\sigma$-algebra. The answer is no. I mention in my comment that $X = \lbrace 0,1\rbrace$ with the topology $\mathcal T = \lbrace \emptyset ,\lbrace 0\rbrace , X\rbrace$ provides a counterexample. Here is the full argument. If $\mathcal T$ was generated by a $\sigma$-algebra, then this $\sigma$-algebra would be some subfamily $\mathcal A \subset \mathcal T$. Being a $\sigma$-algebra, $\mathcal A$ would have to contain $\emptyset$ and $X$. However, $\mathcal A\neq \lbrace \emptyset , X\rbrace$ since this $\mathcal A$ is already a topology. So $\mathcal A$ must be strictly larger than $\lbrace \emptyset ,X\rbrace$, i.e. $\mathcal A$ must be $\lbrace \emptyset , \lbrace 0 \rbrace , X\rbrace = \mathcal T$. But this is not a $\sigma$-algebra since then $\lbrace 0 \rbrace ^c = \lbrace 1\rbrace\not\in\mathcal A$.

Advice. In introductory measure theory, you will probably meet two kinds of measure spaces:

1. Abstract measure spaces, where the $\sigma$-algebra of measurable sets has nothing to do with a topology, since no topology is given.
2. Topological spaces equipped with the Borel $\sigma$-algebra mentioned above. By far the most important example is $\mathbb R$.

Try to get familiar with the first case -- and remember that this case is non-topological! When you have done this, you will naturally view the second case as simply a way to construct a specific $\sigma$-algebra $\mathcal A$ on a topological space $X$ such that $\mathcal A$ encodes some of the topological structure of $X$.

Addendum. Given a measure space $(X ,\mathcal A)$, can we always find a topology $\mathcal T$ on $X$ so that $\mathcal A$ is the Borel $\sigma$-algebra of $\mathcal T$? This is the question in the excellent link given by @drhab in the comments. It turns out that the answer is no (a counterexample is given in the accepted answer). This shows that even theoretically only some $\sigma$-algebras admit a topological interpretation -- that is, include a generating set which is a topology.