Proof of De la Vallée-Poussin's Test Here's what I'm trying to prove.
Suppose that $A \subseteq \mathbb{R}^n$ and $J = [c,+\infty)$. Let $f: A \times J \to \mathbb{R}^p$ be continuous and suppose that there exists a function $g: J \to [0,+\infty)$ such that:
$$\forall (x,t) \in A \times J: \|f(x,t)\| \leq g(t)$$
Suppose that $\int_{c}^{\infty} g(t) \ dt$ is convergent. Then, $\int_{c}^{\infty} f(x,t) \, dt$ is uniformly convergent for $x \in A$.
Here's the definition of uniform convergence (I'm pretty sure that this is really just an extension of the definition of improper integrals on the real line).
Again, let $f$ be as given above. Then, define $F(x) = \int_{c}^{\infty} f(x,t) \, dt$. We say that $f$ is uniformly convergent for $x \in A$ iff:
$$\forall \epsilon > 0: \exists D > c: x \in A \land d \geq D \implies \left\| F(x)-\int_{c}^{d} f(x,t) \right\| < \epsilon$$

Proof Attempt:
Observe that for any fixed $x \in A$, we have:
$$\forall i \in \{1,2,\ldots,p\}: |f_i(x,t)| \leq \left\|f(x,t)\right\| \leq g(t)$$
$$\implies \forall d > c: \int_{c}^{d} |f_i(x,t)| \ dt \leq \int_{c}^{d} g(t) \ dt$$
Then, since $\int_{c}^{d} g(t) \ dt$ converges as $d \to \infty$, it follows that $\int_{c}^{d} |f_i(x,t)| \ dt$ converges as $d \to \infty$ by comparison. Then, this implies that $\int_{c}^{\infty} f_i(x,t) \ dt$ converges and since the integral of each component function of $f$ converges, it follows that $\int_{c}^{\infty} f(x,t) \ dt$ converges. $\Box$
The thing that I'm somewhat worried about is if I'm missing something crucial over here. Like, I'm just wondering if I've overlooked something or if I would actually have to do a more formal argument using the definition directly. I'd appreciate it if someone could have a look at my working.
 A: Using
$$\int_c^d |f_i(x,t)| \, dt \leqslant\int_c^d g(t) \, dt,$$
and the fact that the integrand on the LHS is nonnegative, you can only conclude pointwise convergence, that is for each $x \in A$,
$$\lim_{d \to \infty} \int_c^d |f_i(x,t)| \, dt = \limsup_{d \to \infty}\int_c^d |f_i(x,t)| \, dt \leqslant \lim_{d \to \infty}\int_c^d g(t) \, dt$$
To prove uniform convergence, you need to show that
$$\lim_{d \to \infty} \sup_{X \in A}\left|\int_d^\infty f_i(x,t) \, dt \right| = 0$$
We have,
$$\left|\int_d^R f_i(x,t) \, dt \right| \leqslant\int_d^R |f_i(x,t)| \, dt \leqslant\int_d^R g(t) \, dt \leqslant \int_d^\infty g(t) \, dt$$
Taking the limit as $R \to \infty$ we get
$$\left|\int_d^\infty f_i(x,t) \, dt \right|\leqslant  \int_d^\infty g(t) \, dt, \\\sup_{x \in A}\left|\int_d^\infty f_i(x,t) \, dt \right|\leqslant  \int_d^\infty g(t) \, dt$$
Since the improper integral of $g$ converges, we get as desired
$$\lim_{d \to \infty}\sup_{x \in A}\left|\int_d^\infty f_i(x,t) \, dt \right|=\lim_{d \to \infty}  \int_d^\infty g(t) \, dt = 0$$
