Suppose we want to prove that among some collection of things, at least one
of them has some desirable property. Sometimes the easiest strategy is to
equip the collection of all things with a measure, then show that the set
of things with the desired property has positive measure. Examples of this strategy
appear in many parts of mathematics.

> **What is your favourite example of a proof of this type?**

Here are some examples:

* **The probabilistic method in combinatorics** As I understand it, a
  typical pattern of argument is as follows.  We have a set $X$ and want to
  show that at least one element of $X$ has property $P$.  We choose some
  function $f: X \to \mathbb{N}$ such that $f(x) = 0$ iff $x$ satisfies
  $P$, and we choose a probability measure on $X$.  Then we show that
  with respect to that measure, $\mathbb{E}(f) < 1$.  It follows that
  $f^{-1}\{0\}$ has positive measure, and is therefore nonempty.

* **Real analysis** One example is [Banach's
    proof](http://www.artsci.kyushu-u.ac.jp/~ssaito/eng/maths/Cauchy.pdf)
    that any measurable function $f: \mathbb{R} \to \mathbb{R}$ satisfying
    Cauchy's functional equation $f(x + y) = f(x) + f(y)$ is linear.
    Sketch: it's enough to show that $f$ is continuous at $0$, since then
    it follows from additivity that $f$ is continuous everywhere, which makes
    it easy.  To show continuity at $0$, let $\varepsilon > 0$.  An
    argument using Lusin's theorem shows that for all sufficiently small
    $x$, the set $\{y: |f(x + y) - f(y)| < \varepsilon\}$ has positive
    Lebesgue measure.  In particular, it's nonempty, and additivity then
    gives $|f(x)| < \varepsilon$.

    Another example is the existence of real numbers that are
    [normal](https://en.wikipedia.org/wiki/Normal_number) (i.e. normal to every base).
    It was shown that almost all real numbers have this property
    well before any specific number was shown to be normal.

* **Set theory** Here I take ultrafilters to be the notion of measure, an
    ultrafilter on a set $X$ being a finitely additive $\{0, 1\}$-valued
    probability measure defined on the full $\sigma$-algebra $P(X)$.  Some
    existence proofs work by proving that the subset of elements with the
    desired property has measure $1$ in the ultrafilter, and is therefore nonempty.

    One example is a proof that for every measurable cardinal
    $\kappa$, there exists some inaccessible cardinal strictly smaller than
    it.  Sketch: take a $\kappa$-complete ultrafilter on $\kappa$. Make an inspired choice of function $\kappa \to \{\text{cardinals } <
    \kappa \}$. Push the ultrafilter forwards along this function to give
    an ultrafilter on $\{\text{cardinals } < \kappa\}$. Then prove that the set
    of inaccessible cardinals $< \kappa$ belongs to that ultrafilter ("has
    measure $1$") and conclude that, in particular, it's nonempty.

    (Although it has a similar flavour, I would *not* include in this list the cardinal arithmetic proof of the
    existence of transcendental real numbers, for two reasons.  First,
    there's no measure in sight.  Second -- contrary to
    popular belief -- this argument leads to an *explicit construction*
    of a transcendental number, whereas the other arguments on this list
    do not explicitly construct a thing with the desired properties.)


(Mathematicians being mathematicians, someone will probably observe that
*any* existence proof can be presented as a proof in which the set of things
with the required property has positive measure. Once you've got a thing
with the property, just take the Dirac delta on it.  But obviously I'm
after less trivial examples.)

**PS** I'm aware of the earlier question [On proving that a certain set is
  not empty by proving that it is actually
  large](https://mathoverflow.net/questions/34390). That has some good
  answers, a couple of which could also be answers to my question. But my
  question is specifically focused on *positive measure*, and excludes
  things like the transcendental number argument or the Baire category
  theorem discussed there.