Start with . Having chosen disjoint , put and
The empty set shows , and finite total variation bounds it above. If , choose with ; if , choose . This is the required recursion, including the initial step .
Disjointness and nonnegativity give , hence . Write . For every measurable ,
Thus every measurable subset of has nonpositive measure. Each finite partition of consequently has variation sum exactly , and therefore . It follows that . Finally countable additivity gives

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