# Why is the set $\{x \mid f(x) \not= g(x)\}$ measurable in a topological hausdorff space?

assuming i have a measure space $$(X, \mathscr{M}_X, \mu)$$ and a measurable space $$(Y,\mathscr{M}_Y)$$, two measurable mappings $$f,g:X \to Y$$ and it holds that $$Y$$ is a topological hausdorff space with a countable base. (I think i am supposed to assume that $$\mathscr{M}_Y$$ is the Borel-$$\sigma$$-Algebra generated by the open/closed subsets of Y)

Now i am wonderung: Why exactly is the set $$\{x \in X \mid f(x) \not= g(x)\}$$ measurable? in other words why is $$\{x \in X \mid f(x) \not= g(x)\} \in \mathscr{M}_X$$

I know that the diagonal $$\Delta Y = \{(y,y)\mid y \in Y\}\subset Y \times Y$$ is closed in any hausdorff space.

I also know that since $$\Delta Y$$ is closed in $$Y \times Y$$, its complement $$(Y \times Y)\setminus \Delta Y$$ is open.

However, $$\{(f(x),g(x)\mid x \in X, f(x) \not= g(x) \}$$ is only a subset of $$(Y \times Y)\setminus \Delta Y$$ for which i can't say whether it is open or closed.

The only idea that's left was to try this:

Pick an arbitrary point $$p = (f(x),g(x))$$ for which $$f(x)\not= g(x)$$ in $$Y\times Y$$. Since $$Y$$ is a Hausdorff space, we can find disjoint open neighbourhoods $$U$$ around $$f(x)$$ and $$V$$ around $$g(x)$$ such that $$U \times V$$ is an open neighbourhood of $$(f(x),g(x))$$ with $$U \times V \cap \Delta Y = \emptyset$$

Since $$\{(f(x),g(x)\mid x \in X, f(x) \not= g(x) \}$$ is a union of all such open neighbourhoods $$U\times V$$ of points $$(f(x),g(x))$$ with $$f(x) \not= g(x)$$ it is open in $$Y\times Y$$. (is this correct?)

If the last paragraph would be correct, i'd be finished.

Thank you very much for any of your help.

EDIT: New summary of what i gathered so far with the help of the recent comments and answers:

As pointed out in the comments, i did the mistake to purely focus on $$B:= \{(f(x),g(x)\mid f(x) \not= g(x), x \in X\}$$ and tried to conclude that its preimage is measurable iff $$B$$ is open which was a wrong assumption in the first place.

Therefore, i do not need $$B$$ itself to be open. It's sufficient to show that the union of all open disjoint neighbourhoods $$U\times V$$ of arbitrary points $$f(x),g(x)$$ for which $$f(x) \not= g(x)$$ is an open set; because what matters is its preimage and not the set (image) itself. Let $$B' := \bigcup\{U \times V \mid U, V \in \mathcal{B}\ ;\ U \cap V = \emptyset\}$$

$$B'$$ is obviously an open set, therefore its preimage is mesaurable.

For the preimage of open disjoint neighbourhoods $$U,V$$ of $$f(x)$$ and $$g(x)$$ respectively, it holds that

$$(f\times g)^{-1}(U\times V) = \{x \mid f(x)\in U \wedge g(x) \in V\} = f^{-1}(U)\cap g^{-1}(V)$$ (this is something i actually didnt know, it didn't know the definition of cartesian products of mappings, that's also the reason i didnt understand alex's question right away)

Since $$U\times V$$ is open, $$(f\times g)^{-1}(U \times V)$$ is measurable. Since $$\mathcal{B}$$ is a countable base the set

$$\{x \in X\mid f(x) \not= g(x) \} = \bigcup \{ (f\times g)^{-1}(U\times V) \mid U, V \in \mathcal{B}, U \cap V = \emptyset\} = \bigcup\{ f^{-1}(U)\cap g^{-1}(V)\mid U, V \in \mathcal{B}, U \cap V = \emptyset\}$$ is a countable union of measurable sets and therefore measurable itself.

• There is obviously a typo in the expression $\{(f(x),g(x)\mid x \in X\}$ which occurs at the start of the 5th and 8th paragraphs. For one thing the opening parenthesis remains unclosed. If $\{(f(x),g(x))\mid x \in X\}$ is what was meant, then I don't see why it would necessarily be a subset of $(Y \times Y)\setminus \Delta Y$, since there's nothing in the definition of the set to eliminate those $\ x\$ for which $\ f(x) = g(x)\$. Was it supposed to have been $\{(f(x),g(x))\mid x \in X, f(x) \ne g(x) \}\$ ? Commented Jan 26, 2019 at 11:00
• hello @lonzaleggiera, you're right, that's a typo. thanks for pointing out. i will edit my question immediately.
– Zest
Commented Jan 26, 2019 at 17:28

$$h(x)=(f(x),g(x))$$ is measurable and $$\{x:f(x)\neq g(x)\}=h^{-1}(Y\times Y-\Delta)$$ is measurable since $$\Delta$$ is closed, $$(Y\times Y)-\Delta$$ is open.

• $h$ is measurable strictly follows from $f,g$ both measurable, right? was my proof correct or did i do mistakes along the way?
– Zest
Commented Jan 23, 2019 at 3:39

I think all presented proofs are incomplete (for instance, they don’t use the fact that $$Y$$ has a countable base $$\mathcal B$$). I also assume that $$\mathcal M_X$$ is closed with respect to intersections and countable unions.

Let $$A=\{x\in X|f(x)\ne g(x)\}$$. Let $$x\in A$$ be any point. Since $$f(x)\ne g(x)$$ and $$Y$$ is Hausdorff, there exists disjoint neighborhoods $$U,V\in\mathcal B$$ of $$f(x)$$ and $$g(x)$$, respectively. Thus

$$A=\bigcup \{f^{-1}(U)\cap g^{-1}(V): U,V\in\mathcal B, U\cap V=\varnothing\}$$

is measurable.

• hello alex, thanks for your consideration. Where did you use the fact that $Y$ has a countable base? And why exactly is $A$ measurable in your answer?
– Zest
Commented Jan 26, 2019 at 23:59
• @Zest Since $Y$ has a countable base the union for $A$ is countable. Then $A$ is measurable, because (as I assumed) $\mathcal M_X$ is closed with respect to (finite) intersections and countable unions. Commented Jan 27, 2019 at 0:11
• i see, thank you very much. I might return with questions tomorrow, highly appreciating your help.
– Zest
Commented Jan 27, 2019 at 0:23
• hello alex, i just edited my question and added my own summary of everything i learned with your help. would you mind taking a look and tell me if my proof is correct? it is very detailed but i am on undergruadate level and it helps me sometimes getting a better idea of everything.
– Zest
Commented Jan 27, 2019 at 10:42

In answer to the question asked, about whether the argument in the last paragraph is correct, I don't believe that it is at it stands.

The problem is with the assertion $$"\ \{( \,f(x),g(x)\ )\mid x \in X, f(x) \not= g(x) \}\$$ is a union of all such open neighbourhoods $$\dots\$$ ." If you take the union of all the open neighbourhoods you describe you will certainly get an open set containing $$\ \{( \,f(x),g(x)\ )\mid x \in X,\, f(x) \not= g(x) \}\$$, but I don't see why it should be equal to it, and I believe I can give a counterexample to the general statement that $$\ \{( \,f(x),g(x)\ )\mid x \in X,\, f(x) \not= g(x) \}\$$ must be open under the hypotheses given

Nevertheless, I think Alex Ravsky's proof shows that the basic idea underlying the argument was sound.

Take $$\ X = Y = \mathbb R\$$—the set of real numbers—, and $$\ \mathscr{M}_X ,\mathscr{M}_Y\ \$$ both to be the Borel sets of $$\ \mathbb R\$$. Let $$\ f\left(x\right)= 0\ \mbox{ and }\ g\left(x\right)= x\ \mbox{ for all } x\in\mathbb R\$$. Then $$\ \{( \,f(x),g(x)\ )\mid x \in X,\, f(x) \not= g(x) \}\$$ is the punctured line segment $$\ \left\{\ \left(\,0,x\,\right)\ \mid\ x\ne0\ \right\}\$$ of $$\ \mathbb R^2\$$, which is not an open set.

Comment on proposed new proof: Apart from some probable typos, itemised below, the new revised proof by Zest seems to me to be correct.

The following appear likely to me to be typos:

• $$B' = \bigcup \left\{ U\times V\mid U,V\in B, U\cap V=\emptyset\right\}\ .$$ This should be $$B' = \bigcup \left\{ U\times V\mid U,V\in{\mathcal B}, U\cap V=\emptyset\right\}$$
• I'm not sure what the observation "Since $$\ B\$$ is a countable base $$\ \mathcal B\ \dots\$$" is trying to say. As far as I can see, all that's necessary here is the observation that $$\ \mathcal B\$$ is a countable base.
• A union sign is missing from the last term on the left of the final series of equations. That is, $$\ \left\{\,f^{-1}\left(U\right)\cap g^{-1}\left(V\right)\mid U,V\in{\mathcal B}, U\cap V=\emptyset\right\}\$$ should be $$\ \bigcup\left\{\,f^{-1}\left(U\right)\cap g^{-1}\left(V\right)\mid U,V\in{\mathcal B}, U\cap V=\emptyset\right\}\$$

The nice observation Alex Ravsky and Zest have made here, and which I had missed, is that $$\ \left\{\,x\in X \mid f\left(x\right)\neq g\left(x\right)\,\right\}\$$ is actually equal to $$\ \left(f\times g\right)^{-1}\left(B'\right)\$$ rather than being a strict subset of it. Since this isn't immediately obvious—at least it wasn't to me, although the proof is fairly straightforward once the fact is realised—I'd like to see a brief proof of that fact included in any formal writeup.

• thanks for your help. well, the union of open sets is open, thus if in fact the set $\{(f(x),g(x)) \mid x \in X, f(x) \not= g(x)\}$ is a union of disjoint open neighbourhood, it is open.
– Zest
Commented Jan 26, 2019 at 23:53
• But how did you get that the set $\{( \,f(x),g(x)\ )\mid x \in X, f(x) \not= g(x) \}$ is a union of disjoint open neighbourhoods? If $\ U\$ and $\ V\$ are disjoint open neighbourhoods around $\ f(x)\$ and $\ g(x)\$ respectively, $\ U\$ may well contain an element $\ u\$ that isn't in the image of $\ f\$, and $\ V\$ an element $\ v\$ that isn't in the image of $\ g\$. The point $\ (u, v)\$ will be in any union that includes $\ U\times V\$ but *not* in $\{( \,f(x),g(x)\ )\mid x \in X, f(x) \not= g(x) \}\$, because there is no $\ x \in X\$ with $\ (f(x),g(x)) = (u,v)$ . Commented Jan 27, 2019 at 0:18
• you're right. i didnt consider that. i will think about it again! thanks for the clarification.
– Zest
Commented Jan 27, 2019 at 0:25
• i have another question. is it possible that we don't even need $B := \{ f(x),g(x) \mid x \in X, f(x)\not= g(x)\}$ to be open in $Y$? what matters it that even if $B$ might only be a subset of an open set as you suggested, lets say $B\subset B'$ and $B'$ is the union of all disjoint open neighbourhoods of $f,g$ for which $f\not=g$ then we can still conclude that the preimage of $B'$ is measurable, can't we?
– Zest
Commented Jan 27, 2019 at 9:36
• Yes, I think you've done it. There appear to be some typos in your revised proof, but otherwise it seems to me to be correct. Well done Commented Jan 27, 2019 at 13:33