Any finite measure space can be written as finite disjoint union of measurable set with arbitrarily small measure

Question

$(X,\mathscr{A},\mu)$ be a finite measure space then is the following true ?

For every $\delta>0$, $X$ is finite disjoint union of measurable sets such that the measure of the set(s) are less or equal to $\delta$.

I tried this and could figure out a proof for lebesgue measure on $[0,1]$. For general case I couldn't.

Context: I was trying to prove the converse part of Proposition 4.23 given in the image, which will follow from Lemma 4.22 given there but I couldn't prove the lemma.

Answer 1

False

For example take $X =\{1,2\}$ with each element having measure $1/2$. Then every nonempty set has measure $1/2$ or $1$ and so the claim fails for $\delta=1/4$.