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Let $X$ be a measurable space and $f$ a complex measurable function defined on $X$. Why is the set $E=\{x:f(x)=0\}$ measurable?

I know it is not complicated but I just started studying measurability.

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    What's the definition you're using of complex measurability? – 2017-02-08
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    $f=u+i*v$ where $u$ and $v$ are real measurable functions – 2017-02-08
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    And real-valued measurable functions? Their definition ? Inverse images of what are measurable? – 2017-02-08
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    a real valued function $f : X->Y$ where $X$ is a measurable space (equipped with a sigma-algebra) and $Y$ is a topological space is measurable if $f^{-1}(V)$ is measurable for every $V$ open set of $Y$ – 2017-02-08

3 Answers 3

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OK, following the additions in the comments: write $f = u + iv$, so $u(x) = \operatorname{Re}(f(x))$ and $v(x) = \operatorname{Im}(f(x))$. $u$ and $v$ are then measurable, as $f$ is.

Then $$E = f^{-1}[\{0\}] =u^{-1}[\{0\}] \cap v^{-1}[\{0\}] = \cap_n (u^{-1}(-\frac{1}{n} ,\frac{1}{n})] \cap \cap_n v^{-1}[(-\frac{1}{n}, \frac{1}{n})]$$

which is the countable intersection of measurable sets in the domain, as inverse images of open sets under $u$ or $v$ are measurable. So this is measurable, as measurable sets form a $\sigma$-algebra.

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$ E=f^{-1}\{0\} $ and $\{0\}$ is measurable set with measure $0$. Now since $f$ is a measurable function so it pulls back measurable set to measurable set and hence $E$ is measurable.

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    why is $\{0\}$ a measurable set – 2017-02-08
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    Its finite set, and so its outer measure is zero and hence its measurable. Here measurable means Lebesgue Measure. – 2017-02-08
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    @JonathanBaram A measurable function is always between measurable spaces. You didn't specify it but we all assume that you are dealing with some standard $\sigma$-algebra on $\mathbb{C}$, i.e. at least Borel sets and $\{0\}$ is a Borel set (since it is closed). – 2017-02-08
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    I just started the study of measurability and I am only using basic definitions such as those I wrote in the comments, I did not see Borel sets yet – 2017-02-08
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    Neither Lebesgue measure or even what is a measure – 2017-02-08
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    @JonathanBaram Without any assumption on $\sigma$-algebra on $\mathbb{C}$ this is simply not true. For example with trivial $\sigma$-algebra (i.e. $\{\emptyset, \mathbb{C}\}$) every function $X\to\mathbb{C}$ is measurable. Since any subset can be preimage of $0$ then that would mean that every subset of $X$ is measurable. The implicit assumption here **has to be** that at least $\{0\}$ is measurable in $\mathbb{C}$. – 2017-02-08
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    So why would it not be true? – 2017-02-08
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    @JonathanBaram Because $X$ is arbitrary. Pick for example $X=\mathbb{C}$ with the same trivial $\sigma$-algebra. Pick $f:X\to\mathbb{C}$, $f(x)=x$. Then the preimage of $0$ is $\{0\}$. And this set is not measurable in $X$ because $X$ has trivial $\sigma$-algebra. – 2017-02-08
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    I understand, I guess it had to be implicitly accepted as you pointed out. But still, in the accepted answer we do not use this fact and I cant see something wrong with the proof – 2017-02-08
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    @JonathanBaram Yeah, because the condition $f^{-1}(U)$ is measurable for $U$ open is actually equivalent to the fact that $f$ is Borel-measurable (i.e. measurable with Borel $\sigma$-algebra on $\mathbb{C}$). The function from my previous comment does not have this property. – 2017-02-08
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We have $$E = \{|f| \geq 0\} \cap \{|f| \leq 0\}$$ which is clearly measurable since both $\{|f| \geq 0\}$ and $\{|f| \leq 0\}$ are measurable.

A complex-valued function is measurable if and only if $\operatorname{Re} f$ and $\operatorname{Im} f$ are measurable, i.e. for any $\alpha \in \mathbb{R}$ one and therefore all of the following sets are measurable: $$\{f > \alpha\} \qquad \{f \geq \alpha\}\qquad \{f < \alpha\}\qquad \{f \leq \alpha\}$$

Now we have $$f = \operatorname{Re} f + i\operatorname{Im} f$$

and thus $$|f|^2 = (\operatorname{Re} f)^2 + (\operatorname{Im} f)^2$$ Therefore since $\operatorname{Re} f$ and $\operatorname{Im} f$ are measurable and so is the product and sum of measurable functions we have that $|f|^2$ is measurable which readily implies that also $|f|$ is measurable.

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    How do you show that these two subspaces are measurable – 2017-02-08
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    @JonathanBaram What is your definition of measurability? – 2017-02-08
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    it is in the comments of my question – 2017-02-08
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    I accepted an answer but Im still curious if it is possible to prove the measurability of these sets using only my basic definitions. – 2017-02-08
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    @JonathanBaram No, it is definitely not possible to use your elementary properties for that. – 2017-02-08