Suppose I have a collection of real-valued functions $\{f_n\}_{n\in\mathbb N}$ on a $\sigma$-finite (which we can strengthen to finite if necessary) measure space $(X,\mathcal A,\mu)$ satisfying $f_n\overset{\textrm{a. s.}}{\to}f$ for some $f$, and a real number $\varepsilon>0$ satisfying, for some $1\leq p<\infty$ and all $n\in\mathbb N$, $\lVert f_n\rVert_p\leq\varepsilon$.
So the result I want is $\lVert f_n\rVert_p\to\lVert f\rVert_p$ as $n\to\infty$. I'm pretty sure this is immediate from the continuity of $p$-norms for $1\leq p<\infty$, but I thought I'd try to run through the argument by definition and apply the dominated convergence theorem,
$$\lim_{n\to\infty}\lVert f_n\rVert_p=\lim_{n\to\infty}\left(\int|f_n|^p\,\mathrm d\mu\right)^{1/p}\overset{(\ast)}{=}\left(\int|f|\,\mathrm d\mu\right)^{1/p}=\lVert f\rVert_p,$$
where $(\ast)$ is the application. I first need to find a good function to dominate the integrand uniformly across $n$. The best I could come up with is
$$g(x):=\sum_{k=1}^\infty k\mathbf1_{\left\{k-1<\sup_{n\geq1}|f_n(x)|^p\leq k\right\}}(x),$$
which seems to satisfy $f(x)\leq|g(x)|$ relatively tightly, but then checking for $L^p$ integrability I still can't get better than the form
$$\lVert g\rVert_p^p=\int g^p\,\mathrm d\mu=\sum_{k=1}^\infty k^p\mu\left\{k-1<\sup_{n\geq1}|f_n(x)|^p\leq k\right\}.$$
Am I missing some obvious, better dominating function? Is my constructed $g$ a good dominator, i.e. is it $L^p$?
Best Answer
In general, there need not exist a dominating function which is integrable. Consider, for instance, a moving bump, e.g. $$f_n(x) := 1_{[n,n+1)}(x)$$ on $[0,\infty)$. Then $f_n \to 0$ Lebesgue almost everywhere and $\|f_n\|_{L^1} = 1$ is uniformly bounded in $n$. However,
$$g:=\sup_{n \in \mathbb{N}} f_n = 1_{[1,\infty)}$$
fails to be integrable. Since $g$ is the smallest function which dominates all $f_n$, this means that there does not exist a dominating function which is integrable. Note that this example shows that pointwise convergence $f_n \to f$ and uniform boundedness in $L^p$ does not imply convergence of the $L^p$-norms $\|f_n\|_p \to \|f\|_p$.
Edit: Counterexample in a finite measure space: Consider $(0,1)$ with Lebesgue measure and
$$f_n(x) := n 1_{(0,1/n)}(x).$$
Then $f_n \to f:=0$ pointwise and $\|f_n\|_{L^1} = 1$. However,
$$ \sup_{n \in \mathbb{N}} f_n(x) \geq \sum_{n=1}^{\infty} n 1_{(1/(n+1),1/n)}(x)=:h(x)$$
and
$$\int_{(0,1)} h(x) \, dx = \sum_{n=1}^{\infty} n \int_{1/(n+1)}^{1/n} \, dx \approx \sum_{n=1}^{\infty} \frac{1}{n} = \infty.$$
Consequently, there is no dominating integrable function and $\|f_n\|_{L^1} \to \|f\|_{L^1}$ fails to hold.