= Solution
{c}
Similarly set $g_n(x)=x^{1/2}\mathbf1_{[1/n,1]}(x)$. These nonnegative measurable functions increase to $g$ almost everywhere. The <fundamental theorem of calculus> gives
$$
\int g_n(x)dx=\frac23\left(1-n^{-3/2}\right).
$$
By <monotone convergence>,
$$
\boxed{\int_{\mathbb R}g(x)dx=\frac23.}
$$
The integral is finite and $g\geq0$, so $g$ is <Lebesgue integrable>. Identifying Lebesgue and ordinary integrals of continuous functions on compact intervals justifies each truncated calculation.
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