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I have this exercise on an assignment sheet and I don't really understand what is asked. The formulation is as follows:

Let $X$ be the set of sequences of $n$ zeros and ones ($x\in X$ has the form $x=(x_1,...,x_n)$ where $x_k\in\{0,1\}$) and let $f_k:X\rightarrow\mathbb{R}$ the function $f_k(x)=x_k$. For each $j=1,...,n$, describe explicitly, determining all its sets, the $\sigma$-algebra $A_j$ defined as the smallest $\sigma$-algebra such that all functions $f_k$, $k=1,...,j$ are $A_j$-measurable.

I don't get what I have to do so now I just know that $f_k$ is measurable if and only if $f_k^{-1}[(a,\infty)]$ is measurable. But this set is $\emptyset$ for $a\geq 1$, is $X$ for $a<0$ and is what for $a\in[0,1)$?

Any suggestion is appreciated! Thank you

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1 Answer 1

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If you have a measurable space $(Y, \mathcal Y)$ and some set $X$, and a family of functions $f_i : X \to Y, i \in I$ for some index set $I$, then the $\sigma$-algebra $\sigma(f_i : i \in I)$ generated by these functions on $X$ is the $\sigma$-algebra generated by the inverse images of the measurable sets in $\mathcal Y$ (as by definition these should be contained in $\sigma(f_i : i \in I)$ to make the functions measurable w.r.t. to this $\sigma$-algebra).

In your case the set $X$ is finite, and I suppose $\mathbb R$ is equipped with the Borel $\sigma$-algebra.

Suppose $n = 1$ (the general case is up to you). Then we can write $X = \{0,1\}$ (by indentifying the length $1$ sequenes with the set $X$) and for $f : X \to \mathbb R$ suppose $f(0) = a, f(1) = b$ with $a < b$, then \begin{align*} f^{-1}((-\infty,a-1]) & = \emptyset \\ f^{-1}((-\infty,a]) & = \{0\} \\ f^{-1}((a, b]) & = \{1\} \\ f^{-1}((-\infty, b]) & = \{0,1\} \end{align*} and so we see that all sets must be contained in the generated measure space (as the sets from which I took the inverse images are contained in the Borel $\sigma$-algebra of $\mathbb R$), i.e. the generated measure space equals the discrete $\sigma$-algebra.

Something like that works for arbitrary finite $X$, and in particular for your set of $n$-length sequences of zeros and ones. Hope that was helpful, let me know if you have any further questions!

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  • $\begingroup$ thank you very much, I understood all your explanation. I don't really see how to generalise it now. I mean, in here we only have two elements so we can fix their images. But how do I do that do sequences of $n$ elements? $\endgroup$
    – user194469
    Oct 25, 2015 at 19:30
  • $\begingroup$ Then you have $2^n$ possible values, i.e. $|X| = 2^n$. Despite the fact how you would write it down, do you see what $\sigma$-algebra you get on $X$? $\endgroup$
    – StefanH
    Oct 25, 2015 at 20:05
  • $\begingroup$ Okay, let me add that for your example the functions are called the projection mappings, for example if $n = 2$ then $f_1((0,0)) = f_1((0,1)) = 0$, hence for example $f_1^{-1}((-\infty,0]) = \{ 0 \} \times \{0,1\}$. So you could inspect all the inverse images, now to get a $\sigma$-algebra you just have to compute its closure under intersection, complement and countable union (which for the finite set of your sequences reduces to finite union, and so is equal to computing the boolean algebra containing those sets). $\endgroup$
    – StefanH
    Oct 25, 2015 at 20:23
  • $\begingroup$ Still working on $n=1$. Since $f$ is that projection, we have $f(0)=0$ and $f(1)=1$, right? It's equal to its first component. And why do you compute the preimage of those particular sets? Because they generate $\sigma$-algebra of $\mathbb{R}$? Or because they are enough to prove that the $\sigma$-algebra that we want is the discrete $\sigma$-algebra? $\endgroup$
    – user194469
    Oct 26, 2015 at 15:30
  • $\begingroup$ My intuition would say that that's what will always happen, then. We will always get the discrete $\sigma$-algebra $\endgroup$
    – user194469
    Oct 26, 2015 at 15:33

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