This might be trivial to experts in von Neumann algebra theory, but here goes.
Let $\mu$ be a Radon measure on a locally compact Hausdorff space. Consider the weight $$\varphi: L^\infty(X, \mu) \to [0, \infty]: f \mapsto \int_X f d\mu.$$ Is is true that $\varphi$ is a normal weight on the von Neumann algebra $L^\infty(X,\mu)$?
[It may be possible that we need to impose extra structure on $(X, \mu)$ in order to ensure that $L^\infty(X, \mu)\subseteq B(L^2(X,\mu))$].
For example, if $f_i \nearrow f$ in the strong topology, we need to show that $\varphi(f_i)\to \varphi(f)$ which is some kind of monotone convergence theorem for nets. But I'm not sure such a thing is allowed.
(The topological aspect is irrelevant here; see comment at the end)
Usually you want $(X,\mu)$ to be $\sigma$-finite. When $(X,\mu)$ is $\sigma$-finite, $\varphi$ is normal. Write $X=\bigcup_nX_n$ an increasing union of finite-measure sets. Suppose that $0\leq f_j\nearrow f$ in $L^\infty(X)$. The sot-convergence means that $\|(f-f_j)g\|_2\to0$ for each $g\in L^2(X)$. We consider two cases:
$\varphi(f)<\infty$. Fix $\varepsilon>0$. Choose $n$ such that (by Monotone Convergence) $$ \Big|\varphi(f)-\int_{X_n}f\Big|<\varepsilon. $$ With this $n$ fixed, there exists $j_0$ such that for all $j\geq j_0$ we have $\|(f-f_j)1_{X_n}\|_2^\vphantom2\leq \varepsilon\mu(X_n)^{-1/2}$. Then for $j\geq j_0$ $$ \varphi(f)-\varphi(f_j)\leq \varepsilon+\int_{X_n}f-\int_{X_n}f_j \leq\varepsilon+\|(f-f_j)1_{X_n}\|_2^\vphantom2\,\mu(X_n)^{1/2}<2\varepsilon. $$ Thus $\varphi(f)=\lim_j\varphi(f_j)$.
$\varphi(f)=\infty$. With the same estimate as in the previous case we can fix $n$ and get $j$ such that $$ \int_{X_n}f-\varphi(f_j)\leq \int_{X_n}f-\int_{X_n}f_j \leq\|(f-f_j)1_{X_n}\|_2^\vphantom2\,\mu(X_n)^{1/2}<1, $$ so $$ \varphi(f_j)>-1+\int_{X_n}f $$ As the right-hand-side can be made arbitrarily big and the net $\{\varphi(f_j)\}$ is monotone, we get that $$ \lim_j\varphi(f_j)=\infty=\varphi(f). $$
When $(X,\mu)$ is not $\sigma$-finite, things can get ugly. For instance let $X=\{0,1\}$ and $\mu$ the measure given by $\mu(\{0\})=1$, $\mu(\{1\})=\infty$. Then $$ L^2(X,\mu)=\{g:X\to\mathbb C, \ g(1)=0\},\qquad\qquad L^\infty(X,\mu)=\{g:X\to\mathbb C\}. $$ Put $f=1$, and $f_j=\delta_0$ for all $j$. Then $f_j\leq f$, and for any $g\in L^2(X,\mu)$ we have $$ (f-f_j)(0)g(0)=0g(0)=0,\qquad (f-f_j)(1)g(1)1\,g(1)=0. $$ So $\|(f-f_j)g\|_2=0$, which means that $f_j\nearrow f$ sot. And we have $\varphi(f)=\infty$, and $\varphi(f_j)=1$ for all $j$. So $\varphi$ is not normal.
As for the topology, note that the argument in the $\sigma$-finite case does not use that the measure is Radon in any way. In fact, the topology is not used to define $L^\infty$ nor $\varphi$, so it cannot be expected to play a role.
In the non-$\sigma$-finite example, the measure is not Radon. I'm not entirely sure if the example can be tweaked to get a counterexample with $\mu$ Radon.