Why do we essentially need complete measure space?

While reading the motivation of complete measure space on Wikipedia, what I concluded was, completeness is not really necessary when we define on one measure space and it is necessary when we want to measure on product of measure spaces (is it true ?). If $\lambda$ is measure on $X$ and $Y$ then is it true that $\lambda^2$ is measure of $A$x$B$ and how ? I am not able to understand that $\lambda^2(A\times B)=\lambda(A)\times\lambda(B)$ ? Essentially what is the flaw in the measure without being complete ? Waiting for response. Thanks!

• One can define product measure spaces without completions. But Lebesgue measure is complete, and if we want two-dimensional Lebesgue measure, which is complete, to be be the product of one-dimensional Lebesgue measure with itself, we have to complete the product. Commented May 20, 2012 at 14:06
• By definition of Lebesgue measure we can see that it is complete , right, because subset of a set of measure zero has measure zero ie. the subset is measurable, right ?? But how would that change if we take a product ? Commented May 20, 2012 at 14:20
• If $N$ has measure zero, then $\lambda^2(N\times\mathbb{R})\lambda(N)\cdot\lambda(\mathbb{R})=0\cdot\infty=0$. So if $B$ is any subset of $\mathbb{R}$, then $\lambda^2(N\times B)$ would be $0$ if $\lambda^2$ were complete. But if $B$ is not measurable, the set $N\times B$ is not in the usual product $\sigma$-algebra. But it is in the completion. Commented May 20, 2012 at 14:25
• That a measure space is complete means that every subset of set with measure zero is measurable (and has therefore measure zero too). We know that $\lambda^2(N\times\mathbb{R})=0$ and $N\times B\subseteq N\times\mathbb{R}$ if $B\subseteq\mathbb{R}$. So if the product were complete, $N\times B$ would be in the product $\sigma$-algebra if it were complete. It is not, which is why we complete the product $\sigma$-algebra to get $\lambda^2(N\times B)=0$ for all $B\subseteq\mathbb{R}$. Commented May 20, 2012 at 14:42
• Btw: The Borel $\sigma$-algebra on $\mathbb{R}^2$ does equal the product $\sigma$ of the one-dimensional Borel $\sigma$-algebras! So the motivation Wikipedia gives might not be very convincing. Commented May 20, 2012 at 14:47

5 Answers

We want measure spaces to be complete because we want to treat sets of measure zero as negligible. For example, if two functions $f$ and $g$ satisfy $f(x)=g(x)$ for all $x\in X\setminus N$, and $N$ has measure zero, then we'd like to treat $f$ and $g$ as essentially the same thing. However, without completeness it's possible that $f$ is measurable but $g$ is not.

The issue of completeness is brought into light by the product operation, because the product of complete measures is not always complete. For example, let $A\in [0,1]$ be a nonmeasurable set. The set $A\times \{0\}\subset [0,1]\times [0,1]$ is not measurable with respect to the product measure $\lambda\otimes\lambda$. However, $A\times \{0\}\subset [0,1]\times \{0\}$ and the latter set has product measure $0$. So, once we take the completion of the product measure, $A\times \{0\}$ becomes a measure $0$ set.

• Product measures are complete essentially by construction (they are almost always defined via the Carathéodory construction, which yields complete measure spaces). The point is that their restriction to the product $\sigma$-algebra (the algebra generated by the measurable rectangles) is usually not complete.
– t.b.
Commented May 20, 2012 at 15:17
• @t.b. Well, Folland's "Real Analysis" is an exception to almost always, and since it was my textbook in measure theory...
– user31373
Commented May 20, 2012 at 15:25
• However, without completeness it's possible that f is measurable but g is not. What's wrong with this? Since $f$ and $g$ differ only on a null set, they can be regarded as the same in measure theory.
– JHW
Commented Aug 21, 2015 at 15:13
• because the product of complete measures is not always complete. I don't see why this justifies the completion of a measure. If a product of complete measures is not complete, why do we need the completion of it?
– JHW
Commented Aug 21, 2015 at 15:21

One place in probability theory where complete measures are used is the theory of stochastic processes. We have a stochastic process $X_t$ indexed by reals $t$, so there are uncountably many of them. Certain combinations or these are important, but (as far as can be proved) only equal almost everywhere to a countable combination. With complete sigma-algebra, that is enough for us to conclude that this combination is measurable.

I am giving an answer of this question Philosophically...

All intuition of Measure Theory comes from Probability Theory (finite measure theory ). In Probability certain event is impossible then all its sub events are also impossible (usually).

• In probability theory, it is very common to work with incomplete probability spaces. See for example here. Commented May 20, 2012 at 14:45

I've spent some time on checking whether the completeness of measures plays a role in theorems. At first glance I couldn't find any advantage, so I think some of you maybe interested in it.

There is a so-called projection theorem (Measurable projection theorem proof reference), where completeness is essentially needed.

The second theorem where we assume that the measure is complete (it may be relaxed) is the so-called Scorza-Dragoni theorem.

In fact, the Scorza Dragoni theorem uses projection theorem.

When proving that something holds almost surely, i.e, that its negation never holds (say, the event $A$), it comes handy to 'bound' such an event with a bigger one of measure zero. Then, we do not need to prove that the original event is measurable.

Now, I am sure we can also think of 'practical' cases in which is hard (or not possible) to prove that $A$ is measurable when there is no completeness, while being easy to bound $A$ it by a set of measure zero.