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Suppose in an atomless space, $f\in L^1$ and $||f||_1=1$

I want to prove there exists $g$ and $h$, $g\neq h$, such that $||g||_1=||h||_1=1$ and $f=(1-\lambda)g+\lambda h, 0 \lt \lambda \lt 1$

I'm not sure how to start this question, any tips would be greatly appreciated

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    sorry, you are right. the measure is given to be non-atomic2011-04-02

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There is a subset $E$ of the domain of positive finite measure and a positive number $a$ such that $a\leq|f|$ on $E$. This is true because the sets $\{x:|f(x)|>\frac{1}{n}\}$ must have finite measure for all positive integers $n$, and they can't all have measure $0$. Let $F$ be a subset of $E$ with half the measure of $E$, using the fact that the measure is nonatomic. Define $g$ by $g(x)=f(x)$ off of $E$, $g(x)=f(x)+a\mathrm{sgn}(f(x))$ on $F$, and $g(x)=f(x)-a\mathrm{sgn}(f(x))$ on $E\setminus F$. Define $h$ by $h=2f-g$ (which can be described similarly to $g$ but with sign changes). Then $g\neq h$, $\|g\|_1=\|h\|_1=1$, and $f=\frac{1}{2}(g+h)$.