UMD Analysis Qualifying Exam/Jan09 Real

Change of variable
By change of variable (setting u=nx), we have

$$ \int |f_n(x)| dx = \int \frac{|f(u)|}{n^{\alpha +1}} du \quad \quad (*)\!\,$$

Monotone Convergence Theorem
Define $$ u_n(x) = \sum_{i=1}^{n} |f_i(x)| \!\,$$.

Then, $$ u_n      \!\,$$ is a nonnegative increasing function converging to $$ \sum_{i=1}^{\infty} |f_i(x)| \!\,$$.

Hence, by Monotone Convergence Theorem and $$ (*) \!\,$$

$$ \begin{align}

\int \sum_{i=1}^{\infty} |f_i(x)| dx &= \sum_{i=1}^{\infty} \int |f_i(x)| dx \\

&= \sum_{i=1}^{\infty} \int \frac{|f(x)|}{i^{\alpha +1 }}dx \\ &= \left( \int |f(x)| dx\right) \left( \sum_{i=1}^{\infty} \frac{1}{i^{\alpha +1 }}  \right) \\ &< \infty

\end{align} \!\,$$

where the last inequality follows because the series converges ($$ \alpha >0 \!\,$$ ) and $$   f \in L^1        \!\,$$

Conclusion
Since

$$\int \sum_{i=1}^\infty |f_i(x)| dx < \infty \!\,$$,

we have almost everywhere

$$\sum_{i=1}^\infty |f_i(x)| < \infty \!\,$$

This implies our desired conclusion:

$$\lim_{i \rightarrow \infty} f_i(x) = 0 \quad \mbox{a.e.}\!\,$$