The proof is straightforward if we assume Lebesgue integration
theory: Since $g$ is Riemann integrable, it is bounded. Choose $M>0$
such that $|g(x)|\leq M$ for all $x\in[0,1]$. For each $x\in[0,1)$,
$g(x^{n})\rightarrow g(0)$ as $n\rightarrow\infty$ because $g$
is continuous at $0$. Moreover, $|g(x^{n})|\leq M$. By Dominated
Convergence Theorem, we have
\begin{eqnarray*}
& & \lim_{n\rightarrow \infty}\int_{0}^{1}g(x^{n})dx\\
& = & \int_{0}^{1}\lim_{n\rightarrow\infty}g(x^{n})dx\\
& = & \int_{0}^{1}g(0)dx\\
& = & g(0).
\end{eqnarray*}
Alternative approach that does not invoke Lebesgue integration theory:
Since $g$ is Riemann integrable, it is bounded. Choose $M>0$ such
that $|g(x)|\leq M$ for all $x\in[0,1].$ Let $\varepsilon>0$ be
given. Choose $\delta>0$ such that $|g(x)-g(0)|<\varepsilon$ whenever
$x\in[0,\delta]$. Choose $a\in(0,1)$ such that $a>1-\varepsilon$
. Observe that $a^{n}\rightarrow0$, so there exists $N$ such that $a^{n}\in[0,\delta]$
whenever $n\geq N$. Note that for any $x\in[0,a]$, we have that
$0\leq x^{n}\leq a^{n}$, so $x^{n}\in[0,\delta]$ whenever $n\geq N$
and $x\in[0,a]$. Consider
\begin{eqnarray*}
& & \left|\int_{0}^{1}g(x^{n})dx-g(0)\right|\\
& = & \left|\int_{0}^{1}g(x^{n})dx-\int_{0}^{1}g(0)dx\right|\\
& \leq & \int_{0}^{1}\left|g(x^{n})-g(0)\right|dx\\
& = & \int_{0}^{a}\left|g(x^{n})-g(0)\right|dx+\int_{a}^{1}\left|g(x^{n})-g(0)\right|dx.
\end{eqnarray*}
If $n\geq N$, and $x\in[0,a]$, we have $x^{n}\in[0,\delta]$, so
$|g(x^{n})-g(0)|<\varepsilon.$ It follows that $\int_{0}^{a}\left|g(x^{n})-g(0)\right|dx\leq\int_{0}^{a}\varepsilon dx\leq\varepsilon$.
On the other hand,
\begin{eqnarray*}
& & \int_{a}^{1}\left|g(x^{n})-g(0)\right|dx\\
& \leq & \int_{a}^{1}2Mdx\\
& = & 2M(1-a)\\
& \leq & 2\varepsilon M.
\end{eqnarray*}
That is, $\left|\int_{0}^{1}g(x^{n})dx-g(0)\right|\leq\varepsilon(2M+1)$
whenever $n\geq N$. This shows that $\int_{0}^{1}g(x^{n})dx\rightarrow g(0)$.