What does it mean for a set to have Lebesgue measure zero?

I am studying examples of sets with Lebesgue measure zero (e.g. the Cantor Set) but wanted an intuitive description of what this means rather than a formal definition. Thank you.

Let me quote von Neumann twice:

There's no sense in being precise when you don't even know what you're talking about.

Young man, in mathematics you don't understand things. You just get used to them.

The notion of measure extends the notion of length for intervals, or area for "simple enough shapes" in the plane, or volume of "simple enough shapes" in higher dimensional space. But this extension quickly loses its intuitive meaning.

But don't worry. Working with the definitions for a while causes you to develop a whole new intuition as to what are null sets.

Let me focus on the Lebesgue measure here.

The idea behind a null set starts with something that has no length. Like a singleton. It is an interval of length zero, $[a,a]=\{a\}$. Countable additivity means that actually any countable set is also measure $0$. So even $\Bbb Q$ has measure zero.

But the Cantor set shows that also "large" sets, in the sense that they are equipotent with the real numbers themselves, can be measure zero. In other words, it is not cardinality alone that determines the measure.

Okay. So what does it mean? In the case of the Cantor set, we have a crutch being the fact that it is a closed set. Therefore $[0,1]\setminus C$, the complement of the Cantor set, is open, and therefore is the countable union of intervals. The fact that the Cantor set has null measure is saying that if we look at these intervals in the complement, their lengths add up to $1$.

In the case we consider null sets which are not closed, things can be a bit more difficult, but the idea is that we can approximate the measure as close as we want using closed sets (like the Cantor set), so the measure is the limit of measures of open sets. Meaning, we can find a sequence of intervals $\{I_{n,m}\}_{n,m\in\Bbb N}$ such that for every fixed $n$ the intervals $I_{n,m}$ are pairwise disjoint, and if $x_n=\sum_m I_{n,m}$, then $\lim_n x_n=1$.

Now. From a probability point of view, null sets are sets which represent improbable events. They represent events that are not impossible, but just very unlikely. If you choose a point from $[0,1]$ with uniform distribution, the odds of you choosing any single $x$ is $0$, since almost no $y\in[0,1]$ is $x$ itself. But it is not impossible to choose any $x$, since all of them are valid options.

For this reason we use the Lebesgue (or rather the Borel) measure to determine the probability that the number you chose belongs to some subset of $[0,1]$. Measure zero sets, like the Cantor sets, mean that it is very unlikely to choose an element from them. So unlikely that you can ignore that for "most practical purposes".

So, null sets are the sets we can "safely ignore for the most part", when we try to determine the probability of something; and equally, they are the sets whose length is approximated to $0$ by any possible means (be it from inside using compact sets, or outside using open sets).

The Cantor set shows that measure and cardinality are entirely distinct notion of size when it comes to the real numbers. There are other notions like the notion of "meager sets" or "nowhere dense sets", and there are "fat Cantor sets" with positive measure which are nowhere dense and thus witnessing that these are also distinct ways to measure "largeness".

Practically speaking this means that when you integrate functions such sets don't matter, in the sense that if you modify the domain of integration by a set of measure zero, the integral is unchanged.

It means, in some sense, "negligibly small". One way of thinking about this is via the intuition you already have of probability: if you have measurable sets $A \subseteq B$ in a measure space $(X,\Sigma, \mu)$ with $\mu(B) > 0$, and you throw a dart uniformly at random at $B$, the probability of hitting $A$ is precisely $$ \frac{\mu(A)}{\mu(B)}. $$ In particular, if $\mu(A) = 0$, you miss $A$ with probability 1. In other words, even if you throw (countably) infinitely many times, you'd expect to never once hit $A$.