= Solution
{c}
The <monotone convergence theorem> states that if $0\leq h_n\uparrow h$ pointwise almost everywhere and the functions are <measurable>, then $\int h_n\uparrow\int h$, allowing the value infinity.
Interpret the displayed densities as zero outside $(0,1)$, so no undefined logarithm is evaluated there. Each $f_k$ is nonnegative and <measurable>. With $x=e^{-t}$,
$$
\int_0^1-x^{2k}\log x\,dx
=\int_0^\infty t e^{-(2k+1)t}dt
=\boxed{\frac1{(2k+1)^2}},
$$
where the last equality is <integration by parts>. This is finite, proving <Lebesgue integrability>.
The nonnegative partial sums satisfy, for $0<x<1$,
$$
\sum_{k=0}^Nf_k(x)=-\log x\sum_{k=0}^Nx^{2k}
\ \uparrow\ -\frac{\log x}{1-x^2}=g(x).
$$
They vanish outside the interval as well. Monotone convergence therefore gives
$$
\boxed{\int_{\mathbb R}g(x)dx=\sum_{k=0}^\infty\frac1{(2k+1)^2}.}
$$
The series is finite by comparison with the convergent series $\sum_{j\geq1}j^{-2}$, whose convergence follows from the <integral test>. Since $g\geq0$, this finite integral proves absolute <Lebesgue integrability> as required.
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