Let $(X,\mu )$ be a measure space. Then, $L^2(X):=L^2(X,\mu )$ is a Hilbert space in the usual way and we may view $L^{\infty}(X):=L^{\infty}(X,\mu )$ as a subalgebra of bounded operators on $L^2(X)$ via $L^{\infty}(X)\ni f\mapsto M_f$, where $M_f\colon L^2(X)\rightarrow L^2(X)$ is the multiplication operator defined by $[M_f(g)](x):=f(x)g(x)$.
Regarding $L^{\infty}(X)$ as an algebra of bounded operators in this way, the question can be stated as
How can we describe explicitly convergence in the various operator topologies on $L^{\infty}(X)$?
At least one is relatively easy. For example, $\lambda \mapsto M_{f_{\lambda}}$ converges to $M_f$ in the operator norm topology iff $\lambda \mapsto f_{\lambda}$ converges to $f$ in the $L^{\infty}$ norm. It seems that convergence in measure corresponds to ultra-weak convergence. Furthermore, pointwise almost-everywhere convergence can't correspond to any topology. These are perhaps the three most obvious notions of convergence in $L^{\infty}(X)$, but that leaves many operator topologies unaccounted for. Perhaps most of them just don't have a very nice description?
Update #1: It seems as if the weak operator topology corresponds to the weak-$^*$ topology: $\lambda \mapsto f_{\lambda}\in L^{\infty}(X)$ converges to $f\in L^{\infty}(X)$ in the weak-$^*$ topology iff $\lambda \mapsto \int _Xgf_{\lambda}$ converges to $\int _Xgf$ for all $g\in L^1(X)$, and on the other hand, $\lambda \mapsto M_{f_{\lambda}}$ converges to $M_f$ in the weak operator topology iff $\lambda \mapsto \int _Xgf_{\lambda}h$ converges to $\int _Xgfh$ for all $g,h\in L^2(X)$. As $gh\in L^1(X)$ for $g,h\in L^2(X)$, we obtain the $(\Rightarrow )$ direction immediately. For the $(\Leftarrow )$ write $g=u|g|$ for some Borel function $u$ with $|u|=1$, so that we have $\lambda \mapsto \int _Xgf_{\lambda}=\int _X(u|g|^{1/2})f_{\lambda}|g|^{1/2}$ converges to $\int _X(u|g|^{1/2})f|g|^{1/2}=\int _Xgf$.
Update #2: I have decided to ask this on mathoverflow.