Question

If \(f\) is Lebesgue measurable, then \(|f|\) is Lebesgue measurable.

Answer 1

@eliassaab

Answer 2

Hey :D :D @ParthKohli

Answer 3

?

Answer 4

not sure if this might help

Answer 5

https://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes_ch3.pdf

Answer 6

which part?

Answer 7

3.2 i'm thinking

Answer 8

cause the mod will always yield a positive result

Answer 9

Let g(x)=|x|, then g is measurable. To see that let B be a Borel subset of R then B can be written
\[
B=(-\infty,0[\cap B\cup [0, +\infty(\cap B= B_{-1}\cup B_{+}\\
g^{-1}(B)= B_{-1}\cup B_{+}\\
\]
Hence
\[
g^{-1}(B)
\]
is measurable, then g is measurable.

Answer 10

Let g(x)=|x|, then g is measurable. To see that let B be a Borel subset of R then B can be written\[
B=(-\infty,0[\cap B\cup [0, +\infty(\cap B= B_{-}\cup B_{+}\\
g^{-1}(B)=g^{-1}( B_{-})\cup g^{-1}(B_{+})= (-B_{-})\cup B_{+}\\\\

\]

Hence\[
g^{−1}(B)\]
is measurable, then g is measurable.

Answer 11

Now
\[
|f|= g \circ f
\]
is the composition of two measurable function, so it is measurable

Answer 12

Ok great, I thought of a way last night and would like you to tell me if this also works

\(|f|=f \ , f\ge0\\|f|=-f, f\ge0\)

letting \(g=-f\) , we have \(|f|= f \)^\(g\)
since \(f\) is \(M\)-measurable we know \(-f\) is also, and we know the max of two \(M\)-measurable functions is again \(M\)-measurable, and we are done.

Answer 13

we have proved constant multiplication and max/min properties I used.

Answer 14

and obviously that should say \(|f|=-f,f<0\)

Answer 15

Actually
\[
|f|= f\vee(-f)
\]

Answer 16

oh v is max and ^ is min?

Answer 17

ill associate v with union and ^ with intersection
because the union is usually bigger than the intersection

(this is the inner workings of a dyslexic mathematician)

Answer 18

ty again

Answer 19

YW