A net version of dominated convergence?

Question

Let $G$ be a locally compact Hausdorff Abelian topological group. Let $\mu$ be a Haar measure on $G$, i.e. a regular translation invariant measure. Let $f$ be fixed in $\mathcal{L}^1(G, \mu)$. Define the function from $G$ to $\mathcal{L}^1(G)$ that assigns to each $y$ in $G$ the function that takes $x$ in $G$ to $f(xy^{-1})$. I.e. it assigns the $y$-translate of $f$. Is this a continuous function of $y$ into $\mathcal{L}^1$?

On p.94 3.10 of these notes it is asserted that it is. The usual proof of this sort of thing in cases where $G=\mathbb{R}$, say, is you use density of compactly supported continuous functions, and then dominated convergence. The former survives, but dominated convergence does not work on nets. In fact, a limit of a net of measurable functions need not be measurable.

Answer 1

There simply is no version of dominated convergence which holds for nets, not even if all the functions involved are continuous and compactly supported.

Here's a cautionary example on the unit interval which you can modify to concoct all sorts of counterexamples:

Let $\Lambda$ be the set of finite subsets of $[0,1]$ ordered by inclusion. For each $\lambda \in \Lambda$ choose a continuous function $f_{\lambda}: [0,1] \to [0,1]$ such that $f_{\lambda}(x) = 1$ for all $x \in \lambda$ and $\int f_{\lambda} \leq \frac{1}{2}$. Then $f_{\lambda} \to 1$ everywhere on $[0,1]$, while $\frac{1}{2} \leq \int (1 - f_{\lambda}) \nrightarrow 0$…

In particular, (possible failure of) measurability of the limit is not really the issue.

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That said, try the following approach instead (as suggested by KCd):

1. Show that for $g \in C_c(G)$ the map $y \mapsto g_y$ (where $g_{y}(x) = g(xy^{-1})$) is uniformly continuous as a map $G \to C_c(G)$ where the latter is equipped with the sup-norm. More precisely: for every $\varepsilon \gt 0$ there is a neighborhood $U$ of the identity such that $\lVert g_y - g\rVert_{\infty} \lt \varepsilon$ for all $y \in U$.

2. If $g \in C_{c}(G)$ then there is a compact set $K$ outside of which $g$ vanishes. Take a symmetric compact neighborhood $K'$ of the identity. Then $KK'$ is compact and thus has finite Haar measure. Apply 1. to find a compact neighborhood of the identity with $C \subset K'$ such that $\lVert g_y - g\rVert_\infty \leq \varepsilon/ \mu(KK')$ for all $y \in C$. Note that $g_{y} - g$ vanishes outside $KK'$ so that $$ \lVert g_y - g\rVert_1 = \int \lvert g_y - g\rvert \leq \int_{KK'} \lVert g_y - g\rVert_\infty \leq \varepsilon \quad \text{for all }y \in C. $$

3. Now use that $C_{c}(G)$ is dense in $L^1(G)$. Given $\varepsilon \gt 0$ choose $g \in C_c(G)$ such that $\lVert f-g\rVert_1 \lt \varepsilon$, and note that invariance of Haar measure gives $\lVert f_y - g_y\rVert_1 \lt \varepsilon$. Choose $C$ for $g$ and $\varepsilon \gt 0$ as in point 2.

   For every net $y_\lambda \to 1$ in $G$ we have $y_{\lambda} \in C$ eventually so that $$ \lVert f_{y_\lambda} - f\rVert_1 \leq \lVert f_{y_\lambda} - g_{y_\lambda}\rVert_1 + \lVert g_{y_\lambda} - g\rVert_1 + \lVert g-f\rVert \lt 3\varepsilon $$ and we conclude that $y \mapsto f_y$ is (uniformly) continuous for all $f \in L^1(G)$.

4. Generalize to $L^p(G)$ for $1 \leq p \lt \infty$.

5. Added: To see that the action of $G$ on $L^\infty(G) = L^1(G)^\ast$ is not strongly continuous, consider $G = S^1$ and let $I = [0,t]$ with $t$ be a non-trivial interval. Then the characteristic function $f$ of $I$ has the property that $\lVert gf - f\rVert_\infty = 1$ whenever $g \neq 1$. In particular, for $g_n \to 1 \in G$ we can't have $g_n f \to f$. In fact, it is essentially the definition of right uniform continuity of $f$ that $\lVert gf - f\rVert_\infty \to 0$ whenever $g \to 0$ in $G$.