What is the difference between a null set and an empty set?

[Bipolarity]

I am a bit confused on the difference between the two. Different sources are giving me different results, so I suppose it depends on context. According to some sources, they are the same thing. According to others, the empty set is a set containing no elements, represented by ∅ whereas the null set is any set of measure 0, i.e. having finitely many elements.

My context in asking this question is in proving something about the Laplace transform:
If there is some $a \in ℝ$ for which $\mathcal{L}(f(t)) = \mathcal{L}(g(t))$ on $(a,∞)$, then the set of points t on $[0,∞)$for which $f(t)≠g(t)$ is a null set.

Thanks!

BiP

 

[slider142]

The quote that you used is using the context of a set of measure 0, not a set containing no elements. The same is true for any two functions whose definite integrals over the same interval are equal, which is probably the theorem they use to prove this statement, as it is a simpler statement.

 

[AlephZero]

You may find some old (and/or elementary) books on set theory use "null set" to mean "empty set", but I think the current definitions are that the empty set has no elements, and a null set has measure 0.

Note, a null set can contain an infinite number of elements. Any countable union of null sets is null.

 

[Bipolarity]

You may find some old (and/or elementary) books on set theory use "null set" to mean "empty set", but I think the current definitions are that the empty set has no elements, and a null set has measure 0.

Note, a null set can contain an infinite number of elements. Any countable union of null sets is null.

I see. I am not an expert on measure theory, so to me, measure 0 means finite number of elements, but there's certainly more to it than that I can imagine. What exactly is the meaning of measure 0?

Thanks by the way.

BiP

 

[slider142]

Intuitively, it means that the set contributes a "volume" of 0 to integrals over sets containing that set. Ie., it is invisible to the integral because the set has no length, width, height, etc. But it means a bit more than just "0-dimensional".
While a set containing a finite amount of removable discontinuities does constitute a set of measure 0, that is not the defining usage of the term. The usual definition of the term, at least for the often used Lebesgue measure, is to define it as a type of "mini-integral", the "smallest" collection of sets containing our set of points, where "smallest" is defined by a measure of length. For Lebesgue measure, we measure the length of a closed 1-dimensional real interval [a, b] to be b - a, our intuitive idea of length. So we say l([a, b]) = b - a. We can then build up the idea of n-dimensional volume by defining the volume of an n-dimensional rectangle in the usual product of lengths way.
Continuing in 1-dimension, we then define the measure of a subset of real numbers to be the greatest lower bound, or infimum, of the lengths of all possible unions of intervals that cover that set (the length of a union of intervals would just be the sum of the lengths of intervals, minus any overlap). So we see that we base our idea of measurable on the requirement that a set consist of intervals, which makes intuitive sense. However, as usual, logic leads us to non-intuitive results when applied strictly.
You already have the intuition that isolated points are not intervals and thus should have measure 0. You may also see that countably infinite sets, such as the set of all rational numbers between 0 and 1, would also have measure 0. But there are also uncountably infinite sets, such as the Cantor ternary set, which have measure 0 as well! There are even sets that are not measurable by this definition. Thus, as with most mathematical objects, although it is based on an intuitive idea, in order to verify that a set has measure 0, we have to strictly apply the definition.