Show that $(\log |x|)^2\notin \text{BMO}([-1,1])$. I am trying to show that $u(x)\equiv (\log |x|)^2\notin \text{BMO}([-1, 1])$ by showing that it doesn't satisfy the John-Nirenberg inequality. If $u\in\text{BMO}[-1, 1])$ then this inequality says that there exists a $c_1, c_2>0$ such that for all cubes $Q\subset [-1, 1]$ and for all $t>0$ we have 
\begin{equation}
|\{x\in Q\ |\ |u(x)-u_Q|>t\}|\leq c_1\exp\left(-\frac{c_2t}{|u|_{\ast}}\right)|Q|,
\end{equation} where $|u|_{\ast}$ is the supremum of mean oscillations over all cubes in $[-1, 1]$ and $u_Q$ is the average of $u$ in the cube $Q$.
So I began by assuming that $u\in\text{BMO}([-1, 1])$, and consequently, the above inequality holds for all cubes in $[-1, 1]$ and $t>0$. Then I tried to construct a sequence of cubes $Q_n\subset [-1, 1]$ and a sequence $t_n>0$ such that $t_n\rightarrow \infty$ but 
\begin{equation*}
|x\in Q_n\ |\  |u(x)-u_{Q_n}|>t_n\}|\nrightarrow 0
\end{equation*} to get a contradiction. However, I kept getting stuck.
I would like to know how to obtain a proof using the John-Nirenberg approach and also if my proof below is correct. It does not use the John-Nirenberg inequality but the definition of BMO.
Working
For each $n\in\mathbb{N}$ and $n\geq 3$ define
\begin{equation*}
Q_n\equiv\left[\frac{1}{n}, 1-\frac{1}{n}\right].
\end{equation*} We have that
\begin{align}
u_{Q_n}&=\frac{1}{|Q_n|}\int_{Q_n}[\log x]^2\ \mathrm{d}x\\
&=\frac{n}{n-2}\left(\left[x(\log x)^2\right]^{1-1/n}_{1/n}-\int_{1/n}^{1-1/n}2 \log x\ \mathrm{d}x\right)\\
&=\frac{n}{n-2}\left(\left[x(\log x)^2\right]^{1-1/n}_{1/n}-2\left[x\log x\right]^{1-1/n}_{1/n} +2\left(1-\frac{2}{n}\right) \mathrm{d}x\right)\\
&= \frac{n-1}{n-2}\left(\log\frac{n-1}{n}\right)^2-\frac{1}{(n-2)}[\log n]^2+2\frac{n-1}{n-2}\log n +2
\end{align} and observe that $u_{Q_n}\rightarrow\infty$ as $n\rightarrow\infty$.
Since $u$ is monotone decreasing and continuous function on $Q_n$, there exists a unique element $\tau_n\in Q_n$ such that $u(\tau_n)=u_{Q_n}$, namely $\tau_n\equiv \exp(-\sqrt{u_{Q_n}})$. By construction, we have that $\tau_n\rightarrow 0$ as $n\rightarrow \infty$. 
Next we compute
\begin{align}
\frac{1}{|Q_n|}\int_{Q_n}|u(x)-u_{Q_n}|\ \mathrm{d}x &\geq\frac{n}{n-2}\int_{\tau_n}^{1-\frac{1}{n}}u_{Q_n}-[\log x]^2\ \mathrm{d}x\\
&\geq \int_{\tau_n}^{1-\frac{1}{n}}u_{Q_n}-[\log x]^2\ \mathrm{d}x\\
&=\left(1-\frac{1}{n}-\tau_n\right)u_{Q_n}-\int_{\tau_n}^{1-\frac{1}{n}}[\log x]^2 \ \mathrm{d}x\\
&\equiv \mathcal{I}_n+\mathcal{II}_n.
\end{align}
Similar to earlier calculations, we have
\begin{equation}
\mathcal{II}_n=-\left[x(\log x)^2\right]^{1-1/n}_{\tau_n}+2\left[x\log x\right]^{1-1/n}_{\tau_n} -2\left(1-\frac{1}{n}-\tau_n\right), 
\end{equation}and consequently, $\mathcal{II}_n\rightarrow -2$ as $n\rightarrow\infty$. At the same time $\mathcal{I}_n\rightarrow\infty$ as $n\rightarrow\infty$ so we have that
\begin{equation}
\frac{1}{|Q_n|}\int_{Q_n} |u(x)-u_{Q_n}|\ \mathrm{d}x\rightarrow \infty\Rightarrow u\notin\text{BMO}([-1, 1]).
\end{equation}
 A: I suggest focusing more on estimation than on precise integration. You should not have to integrate $\log^2 x$, or to find $u_Q$ exactly. Estimates are good enough. 
First, a solution using the John-Nirenberg lemma. Fix $Q=[-1,1]$. For $t>2|u_Q|$ we have
$$
\{x\in Q\ |\ |u(x)|>t\} \subset 
\{x\in Q\ |\ |u(x)-u_Q|>t/2\}
$$
If $u$ was in BMO, the measure $|\{x\in Q\ |\ |u(x)|>t\}|$ would decay exponentially in $t$. But in fact
$$|\{x\in Q\ |\ |u(x)|>t\}| = 2\exp(-\sqrt{t})$$
which decays slower. 

Next, an approach directly from the definition. You made some mistakes in integration: in fact, the average of $u$ over $\left[\frac{1}{n}, 1-\frac{1}{n}\right]$ does not tend to infinity; it tends to the average of $u$ on $[0,1]$. The mean oscillation also tends to a finite number, namely the mean oscillation on $[0,1]$. Simply put, you picked a wrong kind of interval as $Q_n$. 
Since the function is integrable, the only way to make the mean oscillation large is to make $Q_n$ small.
I will use the following estimate: 
Claim Let $Q$ be a measurable set. If $c>0$  and $a<b$ are   such that 
$$|\{x\in Q : u(x)\le a\}|\ge c|Q| \quad \text{ and }\quad 
|\{x\in Q : u(x)\ge b\}|\ge c|Q| $$ 
then 
$$
\frac{1}{|Q|}\int_Q |u-u_Q| \ge \frac{(b-a)c}{2}
\tag1 $$
Proof. If $u_Q\le \frac{a+b}{2}$, then 
$$
\int_Q |u-u_Q| \ge \int_{\{x\in Q : u(x)\ge b\}} 
\left(u-\frac{a+b}{2}\right)  \ge c|Q|\frac{b-a}{2} 
$$
Otherwise, 
$$
\int_Q |u-u_Q| \ge \int_{\{x\in Q : u(x)\le a\}} 
\left( \frac{a+b}{2} - u \right) \ge c|Q|\frac{b-a}{2} 
$$
Either way, the claim holds. $\quad \Box$
I would use the above claim with $Q_n = [e^{-n-3},e^{-n}]$,
$a=(n+1)^2$ and $b=(n+2)^2$. Since
$$
|\{x\in Q : u(x)\le a\}| = |e^{-n}-e^{-n-1}|
= \frac{1-e^{-1}}{1-e^{-3}} |Q_n|$$
and 
$$
|\{x\in Q : u(x)\ge a\}| = |e^{-n-2}-e^{-n-3}|
= \frac{e^{-2}-e^{-3}}{1-e^{-3}} |Q_n|$$
we can use $c>0$ independent of $n$. The quantity $b-a$ 
grows indefinitely as $n\to\infty$, which by (1) implies that $u$ is not in BMO. 
