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Problem: Find an example of a positive function $f: [0,1] \to \mathbb{R}_{>0}$ that is of bounded variation, whose reciprocal $1/f$ is integrable but not of bounded variation.

One necessary condition for $f$ is that $\inf_{x \in [0,1]} f(x)=0$, but I don't know how to proceed further.

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    $\begingroup$ Hint: make $1/f$ an integrable function taking all values in $\mathbb{N}-\{0\}$. $\endgroup$ – user10354138 Jul 19 at 4:33
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You have more or less resolved this with the observation that $\inf_{x \in [0,1]} f(x)=0$. Just take the simplest such function, e.g. $$f(x)= \begin{cases}\sqrt{(x)},\qquad x\neq0,\\1,\qquad x=0.\end{cases}$$

Apart from your observation, the only consideration is making sure that the reciprocal is integrable.

Note that any continuous $f$ with the property $\inf_{x \in [0,1]} f(x)=0$, will not be positive.

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