I think the idea of Bill Johnson's counterexample would have been more easily understandable if expressed e.g. as follows: For $i\ge 0$ a natural number, let $U_i=\bigcup_{k=0}^{i}J_{ik}$ where $J_{ik}={]}-1+2\\,k\\,(i+1)^{-1},-1+2\\,k\\,(i+1)^{-1}+(i+1)^{-3}{[}\ $. Putting $U=\bigcup_{i=0}^\infty U_i$ and $K=[-1,1]\setminus U$, then $U$ is open, and $K$ is closed with positive measure since $U$ has measure at most $\sum_{i=0}^\infty\\,(i+1)^{-2}< 1+\int_1^{+\infty}x^{\\,-2}\\,{\rm d\\,}x=2$. Defining $f_i$ on $[-1,1]$ by $t\mapsto 2\\,(i+1)^{\\,2}$ for $t\in U_i$, and $f_i(t)=0$ otherwise, and $f:[-1,1]\owns t\mapsto 1$, it is clear that $\int_{-1}^1(f_i\cdot g)\to\int_{-1}^1(f\cdot g)$ for every continuous $g$. However, this fails if we take as $g$ the characteristic function of $K$ since then $\int_{-1}^1(f_i\cdot g)=0$ but $\int_{-1}^1(f\cdot g)$ equals the measure of $K$.

I think the idea of Bill Johnson's counterexample would have been more easily understandable if expressed e.g. as follows: For $i\ge 0$ a natural number, let $U_i=\bigcup_{\,k=0}^{\,i}J_{\,ik}$ where $J_{\,ik}={\,]\,}{-1}+2\,k\,(i+1)^{-1},-1+2\,k\,(i+1)^{-1}+(i+1)^{-3}{[}\ $. Putting $U=\bigcup_{\,i=0}^{\,\infty}U_i$ and $K=[{-1},1]\setminus U$, then $U$ is open, and $K$ is closed with positive measure since $U$ has measure at most $\sum_{\,i=0}^{\,\infty}\,(i+1)^{-2}< 1+\int_{\,1}^{+\infty}x^{\,-2}\,{\rm d\,}x=2$. Defining $f_i$ on $[-1,1]$ by $t\mapsto 2\,(i+1)^{\,2}$ for $t\in U_i$, and $f_i(t)=0$ otherwise, and $f:[-1,1]\owns t\mapsto 1$, it is clear that $\int_{-1}^1(f_i\cdot g)\to\int_{-1}^1(f\cdot g)$ for every continuous $g$. However, this fails if we take as $g$ the indicator function of $K$ since then $\int_{-1}^1(f_i\cdot g)=0$ but $\int_{-1}^1(f\cdot g)$ equals the measure of $K$.

I think the idea of Bill Johnson's counterexample would have been more easily understandable if expressed e.g. as follows: For $i\ge 0$ a natural number, let $U_i=\bigcup_{k=0}^{i}J_{ik}$ where $J_{ik}={]}-1+2\\,k\\,(i+1)^{-1},-1+2\\,k\\,(i+1)^{-1}+(i+1)^{-3}{[}\ $. Putting $U=\bigcup_{i=0}^\infty U_i$ and $K=[-1,1]\setminus U$, then $U$ is open, and $K$ is closed with positive measure since $U$ has measure at most $\sum_{i=0}^\infty\\,(i+1)^{-2}< 1+\int_1^{+\infty}x^{\\,-2}\\,{\rm d\\,}x=2$. Defining $f_i$ on $[-1,1]$ by $t\mapsto(i+1)^{\\,2}$ for $t\in U_i$, and $f_i(t)=0$ otherwise, and $f:[-1,1]\owns t\mapsto 1$, it is clear that $\int_{-1}^1(f_i\cdot g)\to\int_{-1}^1(f\cdot g)$ for every continuous $g$. However, this fails if we take as $g$ the characteristic function of $K$ since then $\int_{-1}^1(f_i\cdot g)=0$ but $\int_{-1}^1(f\cdot g)$ equals the measure of $K$.

I think the idea of Bill Johnson's counterexample would have been more easily understandable if expressed e.g. as follows: For $i\ge 0$ a natural number, let $U_i=\bigcup_{k=0}^{i}J_{ik}$ where $J_{ik}={]}-1+2\\,k\\,(i+1)^{-1},-1+2\\,k\\,(i+1)^{-1}+(i+1)^{-3}{[}\ $. Putting $U=\bigcup_{i=0}^\infty U_i$ and $K=[-1,1]\setminus U$, then $U$ is open, and $K$ is closed with positive measure since $U$ has measure at most $\sum_{i=0}^\infty\\,(i+1)^{-2}< 1+\int_1^{+\infty}x^{\\,-2}\\,{\rm d\\,}x=2$. Defining $f_i$ on $[-1,1]$ by $t\mapsto 2\\,(i+1)^{\\,2}$ for $t\in U_i$, and $f_i(t)=0$ otherwise, and $f:[-1,1]\owns t\mapsto 1$, it is clear that $\int_{-1}^1(f_i\cdot g)\to\int_{-1}^1(f\cdot g)$ for every continuous $g$. However, this fails if we take as $g$ the characteristic function of $K$ since then $\int_{-1}^1(f_i\cdot g)=0$ but $\int_{-1}^1(f\cdot g)$ equals the measure of $K$.