Because it is possible for a probability density function $p$ to be zero at infinitely many points, and even to be negative at countably many points, there is something that needs proving here. To those intimately familiar with the theory of integration, the result is obvious. My aim in this post is to point out that the obvious is not, however, trivial to show; yet it can be demonstrated using the most basic properties of probabilities and densities with which everyone at any level of learning is conversant.
To justify the first claims above, which might surprise some people, let us note in passing that the density of (say) any positive random variable is zero almost everywhere on the negative numbers, showing densities can have plenty of zeros. We could modify any such density function by setting its values to arbitrary negative numbers at all the integers. This would not change any integral involving the density (even in the sense of Riemann), showing the modified function is also a valid density, despite having many negative values.
To proceed, all we need to know about probability is that
A probability is a countably additive non-negative set function. This is a compact way of stating two of the probability axioms. At the very end we will also need the third axiom, that the total probability is $1,$ to find exactly what a "very high probability" value is.
A density function $p$ for a probability $\Pr$ satisfies $$\Pr(\mathcal E) = \int_\mathcal E p(x)\,\mathrm dx$$ for all sets $\mathcal E$ for which $\Pr$ is defined (the "measurable" sets).
Use your favorite construction for this integral: Riemann, Lebesgue, Kurzweil-Henstock, or whatever, because we will need only the most basic properties of integrals for this analysis. The integral is meant in a general sense: the integral sign $\int$ denotes integrals over spaces of arbitrary dimension. But for simplicity I will refer to the elements of these spaces as "numbers."
For any number $q,$ let $\mathcal E_q$ be the set of numbers $x$ where the density does not exceed $q:$ that is, for which $p(x)\le q.$ The complement $\mathbb R \setminus \mathcal E_0$ therefore is the set of numbers at which $p$ is positive: the set with "very high probability" in the quotation. We need to figure out what its probability could be.
No matter what the number $q$ might be, observe that
$$0 \le \Pr(\mathcal E_{q}) = \int_{\mathcal E_{q}} p(x)\,\mathrm dx \le q\int_{\mathcal E_{q}}\,\mathrm dx .$$
Because the integral on the right cannot be negative, when $q\lt 0$ that integral must be zero if we aren't to arrive at a contradiction. In particular, every set $\mathcal E_{-1/n}$ for whole numbers $n=1,2,3,\ldots$ has zero probability. Let $\mathcal E_{-\omega}$ denote their union: it consists of all numbers where $p$ is negative.
Because the sets $E_{-1/n}$ overlap, it's hard to work with their probabilities. But by writing these sets in order and considering just the numbers newly appearing at each step, they break up into disjoint sets
$$\mathcal E_{-\omega} = \mathcal E_{-1} \cup \mathcal E_{-1/2} \cup \mathcal E_{-1/3} \cup \cdots = \mathcal E_{-1} \cup \left( \mathcal E_{-1/2} \setminus \mathcal E_{-1}\right) \cup \left( \mathcal E_{-1/3} \setminus \mathcal E_{-1/2}\right) \cup \cdots.$$
Writing $\mathcal F_1 = \mathcal E_{-1}$ and $\mathcal F_n = \mathcal E_{-1/n}\setminus \mathcal E_{-1/(n+1)}$ for $n=2, 3, \ldots$ permits us to partition the real numbers as
$$\mathbb R = \left(\mathbb R \setminus \mathcal E_0\right) \cup \left(\mathcal E_0 \setminus \mathcal E_{-\omega}\right) \cup \mathcal F_1 \cup \mathcal F_2 \cap \mathcal F_3 \cup \cdots.$$
(In words: every real number $x$ is one where either $p(x)\gt 0,$ $p(x)=0,$ $p(x) \le -1,$ $-1\lt p(x)\le -1/2, \ldots,$ $-1/n\lt p(x)\le -1/(n+1), \ldots$ and exactly one of these conditions will hold.)
To find the probabilities of each these pieces, note that
Because $\mathcal F_n \subseteq \mathcal E_{-1/n},$ $\Pr(\mathcal F_n) \le \Pr(\mathcal E_{-1/n}) = 0$ for all $n,$ as we have already seen, demonstrating that $\Pr(\mathcal F_n) = 0.$
By construction, the values of $p$ are $0$ or less but greater than any number $-1/n$ on the set $\mathcal E_0 \setminus \mathcal E_{-\omega}.$ Thus, this is precisely the set of zeros of $p.$ Because $p$ is a density, $$\Pr\left(\mathcal E_0 \setminus \mathcal E_{-\omega}\right) = \int_{\mathcal E_0 \setminus \mathcal E_{-\omega}} p(x)\,\mathrm dx = \int_{\mathcal E_0 \setminus \mathcal E_{-\omega}} 0\,\mathrm dx = 0.$$
Applying the countable additivity axiom to this partition gives
$$\begin{aligned}
1 = \Pr(\mathbb R) &= \Pr\left(\mathbb R \setminus \mathcal E_0\right) + \Pr\left(\mathcal E_0 \setminus \mathcal E_{-\omega}\right) + \Pr( \mathcal F_1 ) + \Pr( \mathcal F_2 ) + \Pr( \mathcal F_3 ) + \cdots\\
& = \Pr\left(\mathbb R \setminus \mathcal E_0\right) + 0 + 0 + 0 + 0 + \cdots\\
& = \Pr\left(\mathbb R \setminus \mathcal E_0\right).
\end{aligned}$$
This shows the "very high probability" is necessarily $1.$
