I'm currently studying for an analysis written qual and came across the following question:
Let $m$ denote the Lebesgue measure on $\mathbb R$. Prove that the subset $A$ of $L^1(m)$ defined by $A := \{f \in L^1(m)\ :\ \int_{\mathbb R} |f|dm \leq 1\}$ is closed under pointwise convergence.
I took a topological approach, using the fact that $L^1$ is a complete metric space, under the metric $\rho(f, g) = \int |f - g|dm$. As $A$ is the closed ball of radius 1 centered at $f = 0$, $A$ is also complete. That plus total boundedness implies $A$ is compact. So if $\{f_n\}_{n=1}^\infty$ is a sequence in $A$ that converges pointwise to $f$, it has a subsequence converging in $A$ which implies $f \in A$.
My questions are: 1) is this solution valid? and 2) is there a more measure theoretic approach that works? My first instinct was to use Lebesgue dominated convergence but I couldn't come up with a dominating function.
$A$ is not totally bounded, as the functions $\chi_{(n,n+1)}, n = 1,2,\dots $ are at distance $2$ from one another in the $L^1$ metric.
Hint for the main question: Fatou's Lemma.