Let $p \in [1, \infty]$. Let $(f_n)$ be a Cauchy sequence of probability density functions (p.d.f.) in $L^p (\mathbb R^d)$. Then there is $f \in L^p(\mathbb R^d)$ such that $\|f_n-f\|_{L^p} \to 0$. Convergence in $L^p$ implies pointwise convergence of a sub-sequence. Then $f \ge 0$. I want to verify that $f$ is indeed a p.d.f. It remains to prove that $\int f =1$. I use the following theorem in Brezis' Functional Analysis
Theorem 4.9. Let $\left(f_n\right)$ be a sequence in $L^p$ and let $f \in L^p$ be such that $\left\|f_n-f\right\|_p$ $\rightarrow 0$. Then, there exist a subsequence $\left(f_{n_k}\right)$ and a function $h \in L^p$ such that
- (a) $f_{n_k}(x) \rightarrow f(x)$ a.e. on $\Omega$,
- (b) $\left|f_{n_k}(x)\right| \leq h(x) \quad \forall k$, a.e. on $\Omega$.
Let $(f_{n_k})_k$ be the sequence and $h$ the function given by above theorem. By Fatou lemma, $$ \int \liminf_k f_{n_k} \le \liminf_k \int f_{n_k} \le \limsup_k \int f_{n_k} \le \int \limsup_k f_{n_k}. $$
Then $$ \int f \le 1 \le \int f. $$
This completes the proof.
Of course, the gist of above approach is the existence of the "upper bound" function $h$.
Is there an alternative proof assuming that we don't know Theorem 4.9(b)?
