# Measurable functions on trivial sigma-algebra must be constant?

I'm self-studying measure theory through Axler's "Measure, Integration & Real Analysis" and have come across the following definition of measurable functions:

Suppose $$(X,S)$$ is a measurable space. A function $$f:X\rightarrow \mathbf{R}$$ is called S-measurable if $$f^{-1}(B)\in S$$ for every Borel set $$B\subseteq \mathbf{R}$$.

I'm reading this definition and basically interpreting it in English that every measurable image of the function $$f$$ must also have a measurable pre-image. That is, the Borel sets are measurable, and so for the function to be measurable, the sets that map to the Borel sets via $$f$$ must be $$S$$-measurable. So a measurable function is basically a function that transforms measurable sets into measurable sets in a different space (measurable sets of $$X$$ to Borel measurable sets of $$\mathbf{R}$$ in this case).

It then gives an example: If $$S=\{\emptyset, X\}$$, then the only $$S$$-measurable functions from $$X$$ to $$\mathbf{R}$$ are the constant functions.

And here I'm realizing I must be off. Consider $$X=\mathbf{R}, S=\{\emptyset, \mathbf{R}\}$$. Then $$f:\mathbf{R}\rightarrow\mathbf{R}$$ is measurable if $$f^{-1}(B)\in S$$ for all Borel sets $$B\subseteq \mathbf{R}$$. Consider $$f(x)=x$$. This is clearly a map $$\mathbf{R}\rightarrow\mathbf{R}$$. Consider the Borel set $$(0,1)$$. We then have that $$f^{-1}((0,1))=(0,1)$$. But $$(0,1)\not\in S$$. So wouldn't this mean that it isn't measurable?

Thanks for the help!

• Yes, that means that $f$ isn't measurable (with respect to $S$). To see why the claim is true take an element $x$ from the image of $f$ and look at $f^{-1}(\{x\})$ Feb 7, 2021 at 15:55
• Axler gives the example "If $S=\{\emptyset, X\}$", then the only $S$-measurable functions from $X$ to $\mathbf{R}$ are the constant functions." And now that I'm re-reading it, he says constant functions... not identity functions. Ergo, $f(x)=k$ for some $k\in\mathbf{R}$. $\{k\}$ is a Borel set, and $f^{-1}(\{k\})=X\in S$. Thanks! I guess I do have a basic understanding of the definition. So this shows that any constant functions are $S$-measurable and the identity function isn't $S$-measurable. Nice. Feb 7, 2021 at 16:17

is not correct. It is simply as the definition says: the pre-image of a measurable set has to be measurable. This does not hold for the 'other direction' of the mapping in general. For example take any set $$X$$ and the power set $$\mathcal P(X)$$ as $$\sigma$$-algebra and let $$\mathcal F$$ be a further $$\sigma$$-algebra on $$X$$. Since $$\mathcal P(X)$$ is 'extremly big' every map $$f:X\to X$$ is $$\mathcal P(X)$$-$$\mathcal F$$-measurable. Now consider $$\mathcal F=\{\emptyset, X\}$$. There is no function which transforms all $$\mathcal P(X)$$-measurable sets into $$\mathcal F$$-measurable set if $$X$$ has more than one element.
The example you gave with the $$\sigma$$-algebra $$S$$ and the function $$f=\operatorname{Id}_\mathbb{R}$$ is correct. In this case $$f$$ is not $$S$$-$$\mathcal B$$-measurable. This is the reason why you actually always have to say $$S$$-$$\mathcal B$$-measurable and not only measurable, because beeing measurable depends on both $$\sigma$$-algebras.