1. ## Measurable Function

i couldn't find a way to show this function is measurable.

$\displaystyle f(x)=1$ if $\displaystyle x \epsilon F$ and $\displaystyle f(x)=sin(x)$ if $\displaystyle x\varepsilon [0,1]\F$ where F is the Cantor Set.

i will be appreciate for any help.

edit: i defined a function $\displaystyle g(x)=sin(x)$ over the interval [0,1] but i am not sure about $\displaystyle f \approx g$ because f takes value 1 for some points in [0,1] so there will be discontinuity

2. Originally Posted by gilgames
i couldn't find a way to show this function is measurable.

$\displaystyle f(x)=1$ if $\displaystyle x \epsilon F$ and $\displaystyle f(x)=sin(x)$ if $\displaystyle x\varepsilon [0,1]\F$ where F is the Cantor Set.

i will be appreciate for any help.

edit: i defined a function $\displaystyle g(x)=sin(x)$ over the interval [0,1] but i am not sure about $\displaystyle f \approx g$ because f takes value 1 for some points in [0,1] so there will be discontinuity
The Cantor set has measure 0. Thus the functions f and g are equal almost everywhere. It follows that if one is measurable then so is the other.

3. Write $\displaystyle f(x) = \mathbf{1}_F(x)+\sin x\mathbf{1}_{\left[0,1\right]\setminus F}(x)$. Because $\displaystyle F$ is measurable, $\displaystyle x\mapsto \mathbf{1}_F(x)$ and $\displaystyle \mathbf{1}_{\left[0,1\right]F}(x)$ are measurable. $\displaystyle x\mapsto \sin x$ is measurable because it's continuous.