Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if the Riemann hypothesis is false? The wiki page on Mertens conjecture and the Connection to the Riemann hypothesis says

Using the Mellin inversion theorem we now can express $M$ in terms of 1/ζ as
  $$
 M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\, ds 
$$
  which is valid for $\color{blue}{1} < σ < 2$, and valid for $\color{red}{1/2} < σ < 2$ on the Riemann hypothesis. ... From this it follows that
  $$
    M(x) = O(x^{\color{red}{1/2}+\epsilon}) 
$$
  for all positive ε is equivalent to the Riemann hypothesis, ...

The $\color{red}{\text{red}}$ color indicates the question My question changed, due to anon's comment's, to:


If Riemann was false, would this imply a bound of $M(x)=O(x^{\color{blue}{1}+\epsilon})\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! ---------\;\;$? $ \phantom{------------------------------------------------}$
    Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if RH is false?


A look at Mertens function, makes me think that it should be easy to prove this.

But I still don't have a clue...
 A: There is a trivial bound of $|M(x)| \le x$ for all $x\ge 0$, because the Möbius function is bounded by $1$.  So we already have $M(x) = O(x^1)$ regardless of whether RH is true or false.
This turns out to be rather close to the best known unconditional bound on $M(x)$, which looks like
$$M(x) = O(x \exp( -c\,  \log^{0.6} x \log\log^{-0.2} x))$$
(for instance see this paper of Nathan Ng).  In other words, we do not even have a proven upper bound of the form $O(x^{0.999})$.
Because the Dirichlet series for $\mu(x)$ is just $1/\zeta(s)$, bounds on $M(x)$ can be obtained by Perron's formula using knowledge of the poles of $1/\zeta(s)$, in other words the zeros of $\zeta(s)$.  (It is worth noting that Granville and Soundararajan have recently developed a new approach to many such problems without intimate knowledge of $\zeta(s)$ in the critical strip.)
The fact that we are still rather ignorant about the zeros of $\zeta(s)$ is the reason we don't know a significantly better bound for $M(x)$ than the trivial one.  At the same time, RH being false is not a very strong statement: it just means there is some zero of $\zeta(s)$ with $\Re(s) > 1/2$.  While this does preclude a bound of $M(x) = O_\epsilon(x^{1/2+\epsilon})$, it does not rule out $M(x) = O(x^{3/4+\epsilon})$, if all the zeros of $\zeta(s)$ happen to lie to the left of $\Re(s) = 3/4$.
One last comment: it is somewhat naive to use the observed differences as evidence of how easy it is to prove a bound.  For a more striking example, try graphing the prime gaps function $d(n) = p_{n+1} - p_n$.  You will find that $d(n)$ appears to be $O(\log^2 n)$ (with a pretty small constant), but even assuming RH we don't know how to prove $d(n) = O(\sqrt{n}).$
