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Denote $ f_n = x^\frac{1}{n}f $. We have $ |f_n| \uparrow |f| \in L^1 $, so applying MCT or DCT yields the result

$ \lim_{n \rightarrow \infty} \int_{(0,1)} f_n = \int_{(0,1)} f < \infty $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010