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There is an exercise that I faced in the context of a measure theory and integration course for which I might have a solution, yet I am not entirly sure it's the right one.

Let $B := \{ (0,\infty), (-1,1), (-\frac{1}{2}, \frac{1}{2}), (-\frac{1}{3}, \frac{1}{3}), ... \}$ and let $\mathfrak{A}$ be the $\sigma$-algebra generated by the set $B$.

Now, in the first part I have to decide whether the interval $(-\infty, 0)$ belongs to $\mathfrak{A}$. I would affirm that since $(0, \infty) \in \mathfrak{A}$ and thus, by definition of a $\sigma$-algebra, its complement as well. Is that right?

In the second (and more challenging) part I have to explain whether or not the function $$ f: \mathbb{R} \to \mathbb{R},\, f(x) = \left\{\begin{array}{ll} 1, & \text{falls}\, x > 0 \\ 0, & \text{falls}\, x = 0 \\ -1, & \text{falls}\, x < 0 \end{array}\right. $$

is measurable with respect to $\mathfrak{A}$. The first thing that I don't understand: Is $\mathfrak{A}$ the $\sigma$-algebra for both preimage and image space? Or does it have to be? There is no more information given than the one I gave above.

According to the definition of measurable functions, the preimage $f^{-1}(A) \in \mathfrak{A}$ for every set A in the $\sigma$-algebra of the image space. In case the answer to the first question above is positive, i.e. $\mathfrak{A}$ is also the $\sigma$-algebra of the image space, then I have only to examine whether $f^{-1}(1)$,$f^{-1}(-1)$ and $f^{-1}(0)$ lies in $\mathfrak{A}$. But this is the case for 0 since it is the countably union of a sets in $\mathfrak{A}$ and similarly for -1 and 1. Is that right?

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    For measurability of $f$ it suffices to check that $f^{-1}(-1)=(-\infty,0)$, $f^{-1}(0)=\{0\}$ and $f^{-1}(1) = (0,\infty)$ are all in $\mathfrak{U}$. This isn't as obvious for $0$ as you seem to be implying, as you need to show that $\{0\} \in \mathfrak{U}$ (this was given in the answer below) as well as all the other work up to that point of the question. – 2017-02-14

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Regarding your first question, you have that $(0,+\infty)^c=(-\infty, 0]$ belongs to $\mathfrak{A}$. Now, if you can prove that $\{0\}\in \mathfrak{A}$, then, you'll have that $(-\infty, 0)\in \mathfrak{A}$, because you can write $(-\infty, 0)$ as an intersection of elements in $\mathfrak A$: $$(-\infty, 0)=(-\infty, 0] \cap \{0\}^c$$ But note that $\bigcap_{n\in \mathbb N} (-\frac1n, \frac1n)=\{0\}$, and that means $\{0\}\in\mathfrak A$.

Now, as you say, there's no more information, so we'll suppose that $\mathfrak A$ is the $\sigma$-algebra in both spaces. Now, pick $A\in\mathfrak A$, and study if $f^{-1}(A)\in \mathfrak A$. It is enough to see for the sets on $B$, because $$f^{-1}(\cup_n A_n)=\cup_n f^{-1}(A_n) \qquad f^{-1}(\cap_n A_n)=\cap_n f^{-1}(A_n)$$

So, there are two different kind of sets in $B$: $(0,+\infty)$, and $(-\frac1n, \frac1n)$ for $n\in \mathbb N$. Let's see:

  • $f^{-1}(0,+\infty)=\{x\in\mathbb R: f(x)>0\} = \{x\in\mathbb R: f(x)=1\}=(0,+\infty)\in \mathfrak A$.
  • If $n\in\mathbb N$, $f^{-1}(-\frac1n, \frac1n)=\{x\in\mathbb R: \frac1n
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    Isn't it $(-\infty, 0) = (0, +\infty)^{c}$? – 2017-02-14
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    But why do I have to prove then separately that $\{0\}$ is in $\mathfrak{A}$? – 2017-02-14
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    nice avatar $\;$ – 2017-02-14