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I don't understand the meaning of Measurable function , my lecture told us that $f(x)$ measurable on measurable set $E$ if $E(f>A)=\{x\in X:f(x)>A\}$

$1.\quad$Can you please give me examples for measurable and non-measurable functions?and why?

I tried $f(x)=\sin x ,\quad 1\leqslant x \leqslant \pi$

$A=2:\quad E(\sin x>2)=\emptyset$

$A=1:\quad E(\sin x>2)=\emptyset$

$A=0:\quad E(\sin x>0)=\{x:x\in[0,\pi],\sin x>0\}$

$2.\quad$What can I say about $\sin x$? is it measurable?

Error 404
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  • Reading your question quite some questions arise: is $E$ a set, or some operator on a set (providing some preimage of it)? What is the codomain of function $f$? What is the link with topology? What is the link with Lebesgue-measure? The formal definition provided by the link of your question makes clear that questions like "is this function measurable?" can only be answered if the measurable spaces that are involved are known. So you should mention them in your question. – drhab Jul 18 '16 at 14:08
  • Yes, $E$ is a set, seconde line: "...measurable set $E$" – Error 404 Jul 18 '16 at 14:11
  • @Error404: You can edit your post directly. – parsiad Jul 18 '16 at 14:12
  • In the definition of measurability, a function is measurable with respect to some $\sigma$-field. It doesn't make sense to ask whether a function is measurable simpliciter. – aduh Jul 18 '16 at 14:12
  • Your definition of "$f$ is measurable" is nonsense. You need to get the definition straight before trying to understand it. – David C. Ullrich Jul 18 '16 at 14:13

1 Answers1

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More generally than what you were told,

Let $f\colon X\rightarrow Y$ where $(X,\Sigma_X)$ and $(Y,\Sigma_Y)$ are measure spaces. $f$ is said to be measurable if for each $U\in\Sigma_Y$, $f^{-1}(U)\in\Sigma_X$.

A key point to note is that this is identical to definition of continuity (when replacing the measure spaces by topological spaces). We have simply required that the inverse map preserves measurability, while in topology, the inverse map of a continuous function preserves openness.

Now, let's specialize the definition to the case that you are concerned with. We use the symbol $\mathcal{B}$ to denote the Borel $\sigma$-algebra of $\mathbb{R}$.

Let $f\colon X\rightarrow\mathbb{R}$ where $(X,\Sigma_X)$ and $(\mathbb{R},\mathcal{B})$ are measure spaces. $f$ is said to be measurable if for each $B\in\mathcal{B}$, $f^{-1}(B)\in\Sigma_X$.

The theory of Lebesgue is concerned with being able to give a useful meaning to $\int f d\mu$. The reason we need $f$ to be measurable is because to define integration, we need to be able to approximate $f$ by simple functions.

As for examples of nonmeasurable functions, see this post. However, I recommend that first and foremost you read the approximation of the integral by integrals of simple functions linked to above, since it will give you an idea of why you are doing what you are doing.

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