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

Mathematics Asked by orientablesurface on October 23, 2020

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?)

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

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