To start you are looking to show convergence in the sense of a distribution. So you want $$ \int_{-1}^1 f_n(x) g(x)dx \rightarrow g(0) $$ Now, because integrals are linear and $\int_{-1}^1 f_n(x) dx=1$ we have that $$ g(0)=\int_{-1}^1 f_n(x) g(0) dx. $$ Because you are trying to show convergence you want to look at \begin{eqnarray*} \left|\int_{-1}^1 f_n(x) g(x) dx - g(0)\right|= \left|\int_{-1}^1 f_n(x) (g(x)-g(0)) dx\right| \end{eqnarray*} Separate the integral into three sections ($\int_{-1}^{-c}$, $\int_{-c}^{c}$, $\int_{c}^{1}$) and use the triangle inequality. Here is where you would use the uniform convergence to get rid of the integrals over $[-1,-c]$ and $[c,1]$. All that is left is to show that the integral over $[-c,c]$ is less then some $\varepsilon>0$ arbitrary.