If both $f ge 1$ and $fle 1 $ almost surely then $f = 1_A$, where $A$ is borel

Question

Let $f:[0,1]rightarrow mathbb{R}$ be a positive measurable function. If $fleq 1$ almost surely and $fgeq 1$ almost surely (with respect to lebesgue measure). Show that $f=1_{A}$ for some $Ain B[0,1]$.
I'm not sure how to start.
My attempt is as follows:
So $fgeq 1$ almost surely means there exists some set U in borel algebra on [0,1] such that $mu(U)=0$ on which $f<1$
$fleq 1$ almost surely means that there exists some set V in borel algebra on [0,1] such that $mu(V)=0$ on which $f>1$
By construction of U and V, $Ucap V=varnothing$. Hence, setting $A=Ucup V$, we obtain
However, I have gotten nowhere.
May someone elaborate? What should I modify the problem for the conclusion to hold (if it doesnt hold?)

Kavi Rama Murthy · Answer

There exist sets $E,F$ of measure $0$ such that $f(x) leq 1$ if $x notin E$ and $f(x) geq 1$ if $x notin F$. Let $A=E cup F$. Then $A$ has measure $0$ and $f(x) = 1$ if $x notin A$ (because $x notin A$ implies $x notin E$ and $x notin F$). Now let $D={x: f(x)=1}$. Then $D$ is a Lebesgue measurable set and $f=1_D$ almost everywhere. To complete the proof just note that there is a Borel set $B$ such that $I_D=I_B$ almost everywhere.

Answered by Kavi Rama Murthy on October 23, 2020

If both $f ge 1$ and $fle 1 $ almost surely then $f = 1_A$, where $A$ is borel

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