Since $\sqrt{x}f_n(x)$ converges uniformly to $g(x)$, we have that that for any $\epsilon>0$, there is an $N>0$ so that for $m,n\ge N$, $|\sqrt{x}(f_m(x)-f_n(x))|\le\epsilon$ for all $x\in[0,1]$. Thus, $|f_n(x)|\le|f_m(x)|+\frac{\epsilon}{\sqrt{x}}$. Thus, for $n\ge N$, $f_n$ is dominated by the sum of two integrable functions. Since $f_n(x)\to \frac{g(x)}{\sqrt{x}}$ pointwise, we have that $\displaystyle\lim_{n\to\infty}\int_0^1 f_n(x)\;\mathrm{d}x$ exists by Dominated Convergence.
Furthermore, $\displaystyle\lim_{n\to\infty}\int_0^1 f_n(x)\;\mathrm{d}x=\int_0^1 \frac{g(x)}{\sqrt{x}}\mathrm{d}x$ and $\displaystyle\left\|\frac{g(x)}{\sqrt{x}}\right\|_{L^1[0,1]}\le\left\|f_m\right\|_{L^1[0,1]}+2\epsilon$