I don't think this is silly. For example, if $G=\mathbb R$ then for each $t\in\mathbb R$ consider $$ f_{t,n} = \frac{n}{2} \chi_{[t-1/n,t+1/n]} \in L^1(\mathbb R). $$ Then each $f_{t,n}$ is a unit vector in $L^1(\mathbb R)$. Then given $F\in L^\infty(G)$, we define $$\tilde F(t) = \lim_n \langle F,f_{t,n} \rangle = \lim_n \frac{n}{2} \int_{t-1/n}^{t+1/n} F(s) \ ds$$ Maybe this limit doesn't actually exist-- but the sequence is bounded, so just force it to converge via an ultra-filter limit, or similar. I think you have just described an abstract version of this construction. The point is that $\tilde F$ is little more than a function $\mathbb R\rightarrow\mathbb C$ which is bounded; I don't see why it need have any continuity or measurability properties...

Edit: Actually, maybe a better argument is the following. If you convolve an $L^1(G)$ function by an $L^\infty(G)$ function, then you get a (left or right, depending on taste) uniformly continuous function, which you can then integrate against a bounded measure. So if $(e_\alpha)$ is a bai for $L^1(G)$, then define $T:L^\infty(G)\rightarrow M(G)^*$ by taking an ultrafilter limit: $\langle T(F),\mu\rangle = \lim_\alpha \int_G e_\alpha * F \ d\mu$. This gives the identity on $C_0(G)$ (indeed, on left/right uniformly continuous functions). I think this construction would be well-known to Banach algebraists...

Undeleted: This is perhaps a little more tangential to the original question than I'd hoped. But maybe it gives some hints as to why the conclusions aren't that "worrying"...

I don't think this is silly. For example, if $G=\mathbb R$ then for each $t\in\mathbb R$ consider $$ f_{t,n} = \frac{n}{2} \chi_{[t-1/n,t+1/n]} \in L^1(\mathbb R). $$ Then each $f_{t,n}$ is a unit vector in $L^1(\mathbb R)$. Then given $F\in L^\infty(G)$, we define $$\tilde F(t) = \lim_n \langle F,f_{t,n} \rangle = \lim_n \frac{n}{2} \int_{t-1/n}^{t+1/n} F(s) \ ds$$ Maybe this limit doesn't actually exist-- but the sequence is bounded, so just force it to converge via an ultra-filter limit, or similar. I think you have just described an abstract version of this construction. The point is that $\tilde F$ is little more than a function $\mathbb R\rightarrow\mathbb C$ which is bounded; I don't see why it need have any continuity or measurability properties...

Edit: Actually, maybe a better argument is the following. If you convolve an $L^1(G)$ function by an $L^\infty(G)$ function, then you get a (left or right, depending on taste) uniformly continuous function, which you can then integrate against a bounded measure. So if $(e_\alpha)$ is a bai for $L^1(G)$, then define $T:L^\infty(G)\rightarrow M(G)^*$ by taking an ultrafilter limit: $\langle T(F),\mu\rangle = \lim_\alpha \int_G e_\alpha * F \ d\mu$. This gives the identity on $C_0(G)$ (indeed, on left/right uniformly continuous functions). I think this construction would be well-known to Banach algebraists...

I don't think this is silly. For example, if $G=\mathbb R$ then for each $t\in\mathbb R$ consider $$ f_{t,n} = \frac{n}{2} \chi_{[t-1/n,t+1/n]} \in L^1(\mathbb R). $$ Then each $f_{t,n}$ is a unit vector in $L^1(\mathbb R)$. Then given $F\in L^\infty(G)$, we define $$\tilde F(t) = \lim_n \langle F,f_{t,n} \rangle = \lim_n \frac{n}{2} \int_{t-1/n}^{t+1/n} F(s) \ ds$$ Maybe this limit doesn't actually exist-- but the sequence is bounded, so just force it to converge via an ultra-filter limit, or similar. I think you have just described an abstract version of this construction. The point is that $\tilde F$ is little more than a function $\mathbb R\rightarrow\mathbb C$ which is bounded; I don't see why it need have any continuity or measurability properties...

I don't think this is silly. For example, if $G=\mathbb R$ then for each $t\in\mathbb R$ consider $$ f_{t,n} = \frac{n}{2} \chi_{[t-1/n,t+1/n]} \in L^1(\mathbb R). $$ Then each $f_{t,n}$ is a unit vector in $L^1(\mathbb R)$. Then given $F\in L^\infty(G)$, we define $$\tilde F(t) = \lim_n \langle F,f_{t,n} \rangle = \lim_n \frac{n}{2} \int_{t-1/n}^{t+1/n} F(s) \ ds$$ Maybe this limit doesn't actually exist-- but the sequence is bounded, so just force it to converge via an ultra-filter limit, or similar. I think you have just described an abstract version of this construction. The point is that $\tilde F$ is little more than a function $\mathbb R\rightarrow\mathbb C$ which is bounded; I don't see why it need have any continuity or measurability properties...

Edit: Actually, maybe a better argument is the following. If you convolve an $L^1(G)$ function by an $L^\infty(G)$ function, then you get a (left or right, depending on taste) uniformly continuous function, which you can then integrate against a bounded measure. So if $(e_\alpha)$ is a bai for $L^1(G)$, then define $T:L^\infty(G)\rightarrow M(G)^*$ by taking an ultrafilter limit: $\langle T(F),\mu\rangle = \lim_\alpha \int_G e_\alpha * F \ d\mu$. This gives the identity on $C_0(G)$ (indeed, on left/right uniformly continuous functions). I think this construction would be well-known to Banach algebraists...