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There is an absolute $c>0$ such that, for real $y\ge10$ and every interval $I\subset[2,y]$,

$$\#\{p\in I:p\text{ prime}\} =\int_I\frac{dt}{\log t} +O\!\left(y e^{-c\sqrt{\log y}}\right). \tag{1}$$

The interval may be open, closed, half-open, empty or a singleton.

Proof. For endpoints $2\le a\le b\le y$, subtract the two formulas in external PNT (1). Changing endpoint inclusion changes the prime count by at most two. The integral ignores single endpoints. The PNT error at an endpoint $t\le y$ is bounded by $O(y e^{-c'\sqrt{\log y}})$ with a possibly smaller fixed $c'>0$: the function $t e^{-c\sqrt{\log t}}$ is increasing for all sufficiently large $t$, and the remaining bounded range is absorbed into the constant. The error at $t<10$ is also bounded and is absorbed for $y\ge10$. This proves (1), including degenerate intervals. $\square$

This is a complete deduction from the specified external PNT, not a proof of the PNT itself.

Source. Tao, published paper, published p.799, Lemma 1.6. This page uses that published version.

Bears on. Problem 49.