I saw the following result in a script I am reading about measure theory. But this explanation seems wrong to me on why the indicator function is measurable. For example for the case $a \leq 0$ the set I should get is actually $\Omega \setminus A$, because for $a=0$ given a fixed $A$ should be the elements outside of $A$ which are 0 which lie in $\Omega \setminus A$. Or am I mistaken?

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1 Answer 1


No you are wrong.

For $a\le 0$ ,

$1_A(x) <a\le 0$ implies $1_A(x) <0$

Then there is no element of $\Omega $ such that $1_A(x) <0$ as $1_A(x) =0 \text{ or } 1 $ means $1_A(x)\ge 0$


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