# Two different definitions of General Measurable Function

I've noticed two different kinds of definitions for a Measurable Function.

In Folland's Real Analysis Modern Techniques:

If $(X, \mathcal {M})$ and $(Y, \mathcal {N})$ are measurable spaces, a mapping $f: X \to Y$ is called $(\mathcal {M}, \mathcal {N})$-measurable, or just measurable when $\mathcal {M}$ and $\mathcal {N}$ are understood, if $f^{-1}(E) \in \mathcal {M}$ for all $E \in \mathcal {N}$.

In Halmos' Measure Theory:

Suppose now that in addition to the set $X$ we are given also a $\sigma$-ring $\mathcal {S}$ of subsets of $X$ so that $(X, \mathcal {S})$ is a measurable space. For every real valued (and also for every extended real valued) function $f$ on $X$ we shall write $$N(f) = \{x: f(x) \ne 0 \};$$ if a real valued function $f$ is such that, for every Borel subset $M$ of the real line the set $N(f) \cap f^{-1}(M)$ is measurable, then $f$ is called a measurable function.

I think both kinds of Measurable Function above are talking about a more general edition compared with Lebesgue measurable function. Basically, the difference between them two general measurable functions is whether singling $0$ out of codomain. Why does Halmos do such a removal? Or it is just a evolving process of measure theory from Halmos to Folland coz Folland's book comes out much later?

• I'm no expert, but I suspect this is because Halmos develops measure theory using $\sigma$-rings and Folland using $\sigma$-algebras, and there are some differences to keep in mind between the two theories. See Ramiro Guerreiro's answer to math.stackexchange.com/questions/625602/… for details. Sep 4, 2015 at 21:46
• @GyuEunLee: $\sigma$-ring is more narrow than $\sigma$-algebra? I've noticed the answer in your link that measure on $\sigma$-algebra was extended to $\sigma$-ring. So $\sigma$-ring is much bigger? Sep 4, 2015 at 21:54
• Typically the opposite: a $\sigma$-algebra is always a $\sigma$-ring, but not conversely; a $\sigma$-ring on $X$ need not contain the universal set $X$. But a $\sigma$-ring can be turned into a $\sigma$-algebra by taking the minimal $\sigma$-algebra containing the $\sigma$-ring. Sep 4, 2015 at 21:58
• @GyuEunLee: Ohhh, I see. I appreciate your comment. Sep 4, 2015 at 22:00
• I believe Halmos leaves out $0$ because his purpose is integration of $f$, and where $f = 0$ there are no obstacles to integration. It does not matter if the null set is measurable or not, because it makes no contribution to the integral. Folland is developing a broader theory of functions into other spaces, and therefore has no 0 to play the same role in his development. Sep 4, 2015 at 23:29

Gyu Eun Lee is rigt in his comment. Halmos develops measure theory using σ-rings and Folland using σ-algebras.

Please note that if we limit ourselves to measurable spaces $$(X,\mathcal S)$$ where $$\mathcal S$$ is a $$\sigma$$-algebra, then a real (or extended real) valued function is measurable according to Halmos definition if and only if it measurable according to Foland's definition.

Why Halmos exclude the pre-image of $$0$$? Halmos does it because, if $$(X,\mathcal S)$$ is a measurable space where $$\mathcal S$$ is a $$\sigma$$ -ring but not a $$\sigma$$ -algebra then for any real valued function $$f$$ from $$(X,\mathcal S)$$, $$f^{-1}(\mathbb{R})$$ is not measurable. So, if the pre-image of $$0$$ were not excluded there would NOT be ANY measurable function from $$(X,\mathcal S)$$.

The key point is that Halmos develops the notion of measurable and integrable functions using $$\sigma$$-rings, which is a slightly broader approach than using only $$\sigma$$-algebras, but it makes necessary to deal with additional complexity.

Please also note that in Halmos $$\S$$39, Halmos defines measurable TRANSFORMATION in the same way as Folland (but always considering $$\sigma$$-rings) and explains the "inconsistence" between his definition of measurable function and his definition of measurable transformation.

As I wrote to answer another question some time ago:

Measure Theory using $$\sigma$$-rings will lead to a more complex notion of measurable function, with some non-intuitive results.

Let $$\Omega$$ be a set and let $$\Sigma$$ be a $$\sigma$$-algebra. Let $$f$$ be a function from $$\Omega$$ to $$\mathbb{R}$$. We say that $$f$$ is measurable if for every Borel set $$B$$ in $$\mathbb{R}$$, $$f^{-1}(B)\in \Sigma$$.

Now, suppose that $$\Sigma$$ is a $$\sigma$$-ring and we try to use the same definition. Then, since $$\Omega=f^{-1}(\mathbb{R})$$, either $$\Sigma$$ is a $$\sigma$$-algebra or there will be no measurable function from $$(\Omega,\Sigma)$$.

So, when working with $$\sigma$$-rings, we need a slightly different definition (as we find in Halmos' book). We say that $$f$$ is measurable if for every Borel set $$B$$ in $$\mathbb{R}$$, $$[f\neq 0]\cap f^{-1}(B)\in \Sigma$$.

This second definition allows the existence of measurable functions even if $$\Sigma$$ is just a $$\sigma$$-ring and not a $$\sigma$$-algebra. However, it leads to a few non-intuitive results. For instance: assume $$\Sigma$$ is just a $$\sigma$$-ring and not a $$\sigma$$-algebra. Then any non-zero constant function is NOT mensurable. As a consequence, if $$f$$ is measurable, then it is easy to prove, for instance, that $$f+1$$ is NOT measurable.

So, the theory of measurable and integrable functions is more naturally developed by using $$\sigma$$-algebras, instead of just $$\sigma$$-rings.

• Learned a lot. Will upvote. Sep 5, 2015 at 16:34