Show $g(\mathbf{x}) \leq h(\mathbf{x})$ implies $\int g(\mathbf{x})\mathrm{d}\mathbf{x} \leq \int h(\mathbf{x})\mathrm{d}\mathbf{x}$ Suppose I have $g$ and $h$ from $\mathbb{R}^p\to\mathbb{R}$ such that for all $\mathbf{x}$, $g(\mathbf{x}) \leq h(\mathbf{x})$. I want to prove that the integral over all $\mathbb{R}^p$ of $g$ is less than or equal to that of $h$ (given that the integral over all $\mathbb{R}^p$ of $h$ is finite).
In the 1 dimensional case I could write the infinite integral as a limit of a definite integral and prove that the fact holds for the definite integral over $[a,b]$.
How would I go about proving this for the general case above? I only have a basic understanding of multivariable calculus.

Let $B(0;k)$ denote a ball in $\mathbb{R}^d$ centered at origin with radius k. Once we have shown that
$$ \int_{B(0;k)}h(x) - g(x) dx \geq 0 \quad \forall k \in \mathbb{N}$$
We can then take limit as $k \to \infty$:
$$ \displaystyle\int_{\mathbb{R}^d} h(x) - g(x) dx = \displaystyle\lim _{k \to \infty} \int_{B(0;k)} h(x) - g(x) dx \geq 0$$
