The prime number theorem states that
\[
\pi(x) \sim \frac{x}{\log x}.
\] It is equivalent to $\psi(x) \sim x$ or $\theta(x) \sim x$. Let $M(x) = \sum_{n \leq x} \mu(n)$. In this article we will show that prime number theorem is also equivalent to the estimate $M(x) = o(x)$. We will do so by showing that $M(x) = o(x)$ if and only if $\psi(x) \sim x$. The proof of this equivalence is not entirely obvious.
First we define an associated sum
\[
L(x) = \sum_{n \leq x} \mu(n) \log n.
\] It follows by Abel’s summation by parts formula that
\[
L(x) = M(x) \log x- \int_{1}^{x} \frac{M(u)}{u} \, du.
\] Now using the trivial estimate $M(x) \ll x$ we have
\[
L(x) = M(x) \log x + O(x)
\] which we rewrite as
\[
\frac{L(x)}{x \log x} = \frac{M(x)}{x} + O \left( \frac{1}{\log x} \right).
\] Thus we have $M(x) = o(x)$ if and only if $L(x) = o(x \log x)$. Note that
\[
\Lambda(n) = -\sum_{d \, | \, n} \mu(d) \log d.
\] It follows by Möbius inversion that
\[
\mu(n) \log n = – \sum_{qd = n} \mu(d) \Lambda (q).
\] Using this we write $L(x)$ as
\[
L(x) = -\sum_{n \leq x} \sum_{qd = n} \mu(d) \Lambda(q) = -\sum_{qd \leq x} \mu(d) \Lambda(q) = -\sum_{d \leq x} \mu(d) \psi \left( \frac{x}{d} \right).
\] Now suppose that $\psi(x) \sim x$. Let us write $\psi(x) = x(1 + \delta(x))$, where $\delta(x) = o(x)$. We now take $1 < y < x$ to be some quantity that will depends on $x$ and the choice of which will be made later. We split the sum for $L(x)$ as
\[
L(x) = -\sum_{n \leq y} \mu(n) \psi \left( \frac{x}{d} \right) -\sum_{y < n \leq x} \mu(n) \psi \left( \frac{x}{d} \right)
\] and bound each term separately. First note that
\[\begin{align} \left|\sum_{n \leq y} \mu(n) \psi \left( \frac{x}{n} \right) \right| &= \left|\sum_{n \leq y} \mu(n) \left( \frac{x}{n} + \psi \left( \frac{x}{n} \right) -\frac{x}{n} \right) \right| \\ &\leq x \left|\sum_{n \leq y} \frac{\mu(n)}{n}\right| + \sum_{n \leq y} \left| \psi \left( \frac{x}{n} \right) -\frac{x}{n} \right|. \end{align}
\] Let $\delta^+(x) = \sup_{u \geq x} |\delta(u)|$. Note that $\delta^+(x) = o(1)$. Due to corollary in this post it follows that
\[
\sum_{n \leq y} \mu(n) \psi \left( \frac{x}{n} \right) \ll x + x\sum_{n \leq y} \frac{1}{n} \left| \delta \left( \frac{x}{n} \right) \right| \ll x + x \delta^+ \left( \frac{x}{y} \right) \log y .
\] For the other sum we observe that
\[
\sum_{y < n \leq x} \mu(n) \psi \left( \frac{x}{n} \right) \ll x \psi \left( \frac{x}{y} \right).
\] Taking $y = x/ \log \log x$ we see that
\[
\sum_{n \leq y} \mu(n) \psi \left( \frac{x}{n} \right) \ll x + x \log x \delta^{+}(\log \log x).
\] which implies that
\[
\sum_{n \leq y} \mu(n) \psi \left( \frac{x}{n} \right) = o(x \log x)
\] as $\delta^+ (\log \log x) = o(1)$. For the other sum using the estimate $\psi(x) \ll x$ which follows from assumption we find that
\[
\sum_{y < n \leq x} \mu(n) \psi \left( \frac{x}{n} \right) \ll x \log \log x
\] and so this sum is also $o(x \log x)$. Hence, we have $L(x) = o(x \log x)$.
Now suppose that $M(x) = o(x)$. Note that
\[
\psi(x)- \lfloor x \rfloor + 2 \gamma = \sum_{n \leq x}( \Lambda(n) -1(n) + 2 \gamma e(n) ).
\] Using the identities $\Lambda = \mu * \log$, $1 = \mu * d$, and $e = 1 * \mu$ we get
\[\begin{align} \psi(x)-\lfloor x \rfloor + 2 \gamma &= \sum_{n \leq x} \sum_{qd = n} \mu(d) (\log q -d(q) + 2 \gamma ) \\ &= \sum_{dq \leq x} \mu(d) \left( \log q -d(q) + 2 \gamma \right) \\ &= \sum_{q \leq x} M \left( \frac{x}{q} \right) (\log q -d(q) + 2 \gamma). \end{align}
\] We write $M(x) = x \epsilon(x)$, where $\epsilon (x) = o(x)$. Again we take $1 < y < x$ to be some quantity depending on $x$ which will be chosen later. We now split the sum as
\[ \psi(x) -\lfloor x \rfloor + 2 \gamma = \sum_{q \leq y} M \left( \frac{x}{q} \right) (\log q- d(q) + 2 \gamma) + \sum_{\scriptstyle d \leq x/y \atop \scriptstyle y < q \leq x/d } \mu(d) (\log q -d(q) + 2\gamma).
\] Note that we can bound the first sum as
\[ \left| \sum_{q \leq y} M \left( \frac{x}{q} \right) (\log q- d(q) + 2 \gamma) \right| \leq x\sum_{q \leq y} \frac{1}{q} \left|\epsilon \left( \frac{x}{q} \right)\right| |\log q -d(q) + 2 \gamma|. \] Let $\epsilon^+ (x) = \sup_{u \geq x} |\epsilon(u)|$. Note that $\epsilon^+ (x) = o(1)$. It then follows that the first sum is
\[ \ll x y \log y \, \epsilon^+ \left( \frac{x}{y} \right) \ll xy^2 \, \epsilon^+ \left( \frac{x}{y} \right)
\] Let
\[\Delta(x) = \sum_{n \leq x} (d(n)- \log n + 2 \gamma).
\] Observe that for the other sum we have
\[\begin{align} \left| \sum_{\scriptstyle d \leq x/y \atop \scriptstyle y < q \leq x/d } \mu(d) (\log q -d(q) + 2\gamma) \right| &\leq \sum_{d \leq x/y} \left| \sum_{y < q \leq x/d} \log q- d(q) + 2 \gamma \right| \\ &= \sum_{d \leq x/y} |\Delta(x/d) -\Delta(y)| \\ &\ll \sum_{d \leq x/y} \left( \frac{\sqrt{x}}{\sqrt{d}} + \sqrt{y} \right) \ll \frac{x}{\sqrt{y}}.
\end{align}\] It is easy to see that $\epsilon^+$ is a decreasing function and $\epsilon^+(x) > 0$ for every $x \geq 1$ for if $\epsilon^+(M) = 0$ for some $M \geq 1$, then $\epsilon(x) = 0$ eventually which in turn implies that $\mu(n) = 0$ eventually, a contradiction. We now take $y = \min \left( \sqrt{x}, \epsilon^+(\sqrt{x})^{-1/3} \right)$ and note that
\[
xy^2 \, \epsilon^+ \left( \frac{x}{y} \right) \leq x y^2 \, \epsilon^+(\sqrt{x}) = x \, \epsilon^+(\sqrt{x})^{1/3} = o(x).
\] Since $\epsilon^+(\sqrt{x}) = o(1)$ we obtain that $y \to \infty$ as $x \to \infty$ and so $x/\sqrt{y} = o(x)$. Thus we have shown that
\[
\psi(x) -\lfloor x \rfloor + 2 \gamma = o(x),
\] which implies $\psi(x) \sim x$.