Measurable Function

• May 22nd 2011, 10:05 AM
gilgames
Measurable Function
i couldn't find a way to show this function is measurable.

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

i will be appreciate for any help.

edit: i defined a function $g(x)=sin(x)$ over the interval [0,1] but i am not sure about $f \approx g$ because f takes value 1 for some points in [0,1] so there will be discontinuity
• May 22nd 2011, 11:16 AM
Opalg
Quote:

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

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

i will be appreciate for any help.

edit: i defined a function $g(x)=sin(x)$ over the interval [0,1] but i am not sure about $f \approx g$ 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.
• May 22nd 2011, 12:12 PM
girdav
Write $f(x) = \mathbf{1}_F(x)+\sin x\mathbf{1}_{\left[0,1\right]\setminus F}(x)$. Because $F$ is measurable, $x\mapsto \mathbf{1}_F(x)$ and $\mathbf{1}_{\left[0,1\right]F}(x)$ are measurable. $x\mapsto \sin x$ is measurable because it's continuous.