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If I define a measurable function $f:E \to \Bbb R \cup \pm\{\infty\}$ where $E \subset \Bbb R^d$ is a measurable set as if $\forall a \in \Bbb R$, the set

$$f^{-1}([-\infty,a))=\{x \in E :f(x) < a\}$$ is measurable. Under this assumption is it true that inverse image of a borel set under a measurable function is borel set ? I think not we can only say it is measurable.

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    $\begingroup$ You should have $f: E \rightarrow \mathbb{R} \cup \{ \pm \infty \}$ since you take the inverse image of an interval that is contained only in the extended reals. Or, you can leave the codomain of $f$ alone, and rewrite your statement as $$f^{-1}( (-\infty,a)) = \{x \in E : f(x) < a \}$$ $\endgroup$ – mlg4080 Feb 12 '14 at 7:16
  • $\begingroup$ @mlg4080: sorry I forgot to mention about extended real number system. $\endgroup$ – aaaaaa Feb 12 '14 at 12:46
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It depends on the topology of $E$. Here is a simple example showing $f$ might not be Borel.

By definition, a function is Borel-measurable if the preimage of every open set is Borel. However, it can be shown that it suffices to show the the preimage of every half ray $[-\infty, a )$ is Borel.

Now the exampple. Set $E = \{0,1,2\}$ where the probability (measure) of each point is $1/3$. Now let $f(x)=x$. It is measurable since every subset of $E$ is measurable (this applies also to your definition).

Equip $E$ with the indiscrete topology. That is, $E$ and $\emptyset$ are the only open sets. This implies that the Borel $\sigma$-algebra is $\{\emptyset,E\}$.

Consider $f^{-1}([-\infty,2))=\{0,1\}$. It is the inverse image of a open set but $\{0,1\}$ is not a borel set of $E$. Hence $f$ is not a Borel-measurable function.

Lebesgue gave a characterization of all Borel functions in terms of pointwise convergence and continuity. But it requires a basic knowledge of the Borel hierarchy. Roughly speaking, functions that can be obtained : (1) initially, as the pointwise limit of continous functions (call this class 1) and (2) as the pointwise limit of functions of the previous class (there are called Baire classes). have a tight relationship to each level of the Borel hierarchy. Their preimages lie at each level.

If you are interested in these topic, look at A. Kechris's book Classical Descriptive Set Theory, which is the traditional discipline where Borel sets are studied in depth.

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  • $\begingroup$ Forgive the newbie question here, but with the $E$ you've set up, and the indiscrete topology, would that induce a particular measure? $\endgroup$ – Adrian Keister Apr 2 '14 at 12:27
  • $\begingroup$ Also, it is legal to form the expression $f^{-1}([-\infty,2))$, when the range of $f$ is $E$? $\endgroup$ – Adrian Keister Apr 2 '14 at 12:34
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    $\begingroup$ @AdrianKeister (1) The topology of a set and a measure on it are independent concepts. You can change the topology without changing a measure. There is no obvious "induced measure" in this case although we have the following. Open sets generate a $\sigma$-algebra known as the Borel sets; for the indiscrete topology, the $\sigma$-algebra is just $\{\emptyset,E\}$ and one obtains a measure $\mu(\emptyset)=0$ and $\mu(E)=1$. Other subsets won't be measurable. $\endgroup$ – Mauricio G Tec Apr 2 '14 at 22:53
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    $\begingroup$ (2) The expression is legal since $f$ has counterdomain $\mathbb{R}$ (or the extended system). To define a function, its range must not necessarily equal its counterdomain. $\endgroup$ – Mauricio G Tec Apr 2 '14 at 22:55
  • $\begingroup$ Wonderful! Many thanks. $\endgroup$ – Adrian Keister Apr 3 '14 at 2:38
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I am not sure that this is what you want, it depends on your definition of a measurable function. I assume that you mean that a function is measurable iff the inverse image of every set $([\infty,a)$ is lebesgue measurable. This is equivalent to requiring that the inverse image of every borel set is lebesgue measurable (since the intervals $[-\infty,a)$ generate the borel sigma algebra on $\mathbb{R\cup \{±\infty\}}$.

Here is an counterexample showing that if a function is lebesgue measurable, then the inverse image of every borel set is not always a borel set. A neater way to formulate this is: The class of lebesgue measurable functions is not a subset of the class of borel measurable functions. (The converse is obviously true).

Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}\cup \{±\infty\}$. Set $f(0,y)=1$ for all $y$ belonging to V, where V is the vitaliset. Set $f(x,y)=0$ otherwise.

  • $f$ is lebesgue measurable (I leave it to you to check that. Hint: All you have to know is that the lebesgue algebra is complete).

  • Further, $f^{-1}(1)=\{0\}\times V$, but this set is clearly not borel measurable as that would imply that $V$ is borel measurable. (This follows by the fact that the $x$-section of every borel set in $\mathbb{R}^2$ is measurable in $\mathbb{R}$)

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If we have a measurable function $f: E \rightarrow \mathbb{R}$ all we know is that $\forall V \subset \mathbb{R}$ such that $V$ is open, then $f^{-1}(V)$ is a measurable set. We do not know if $f^{-1}(V)$ is Borel even in the case that $V$ is open.

A function where the inverse image of an open set is Borel, instead of just measurable, is called a Borel mapping, Borel function, or Borel measurable function. Continuous functions, lower semi-continuous, and upper semi-continuous functions are examples of Borel functions.

Here is more information on Borel functions.

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