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I was studying a paper, I had problem understanding the proof of a proposition, First notice that T is a measurable mapping from $([0, 1],B([0, 1]), μ)$ into itself, where μ is a T - invariant probability measure with support $[0, 1]$ and density $f$ with respect to (w.r.t.) the Lebesgue measure m on $([0, 1],B([0, 1])$. Throughout the paper, we assume that any non-empty open interval in $[0, 1]$ contains a non-empty open interval on which $f$ is bounded.We also assume that $T$ admits a finite number of discontinuity points as well as a finite number of local extrema. We denote by $\mathcal{P}$ the set $\lbrace 0 = a_0 < a_1 < \cdots < a_l < a_{l+1} = 1\rbrace ⊂ [0, 1]$ $(l \geq 0)$ of discontinuity points and local extrema of $T$ . By the very definition of $\mathcal{P}$, the transformation $T$ is continuous and monotone on each interval $I_j =]a_j, a_{j+1}[$ for all $j \in \lbrace 0,\cdots, l\rbrace$. As a matter of fact, $T$ is strictly monotone on each interval $I_j$.

I was trying to understand the proof of a proposition, that is this:

If $λ>0$ $μ-a.s$., then:\begin{align*} \max_{j\in \lbrace 0 , 1 ,\cdots ,\mathit{l}_{n}\rbrace} \big(diam(I_{j}^{n})\big) \to 0 \quad as \quad n \to \infty \end{align*}

The proof is this:

Let us assume that \begin{align*} \limsup_{n \to \infty} \max_{j\in \lbrace 0 , 1 ,\cdots ,\mathit{l}_{n}\rbrace}\big(diam(I_{j}^{n})\big) > 0 \end{align*} If this is the case, there exists $ε > 0$ and an increasing sequence $(n_p)_{p\geq 0}$ such that for all $p \geq 0$, $\exists j_{p} \in \lbrace 0, \cdots, l_{n_{p}} \rbrace$ with diam $ diam(I_{j_{p}}^{n_{p}}) \geq \varepsilon$ . By compactness of $[0, 1]$, we can assume without loss of generality that there exists a non-empty open interval $J$ such that for all $p \geq 0$, $ I_{j_{p}}^{n_{p}} \supset J $ . Now, according to Birkhoff’s ergodic theorem, \begin{align*} \frac{1}{n_{p}} \int_{J} \log \vert (T^{n_{p}})^{\prime} \vert d\mu \to \int_{J} \lambda d\mu \quad as \quad p \to \infty \end{align*} By Jensen’s concave inequality and the fact that $μ(J ) > 0$ (since $J$ is a non-empty open interval and the support of $μ$ is $[0, 1]$), \begin{align} \liminf_{p \to \infty} \frac{1}{n_{p}} \log \int_{J} \vert (T^{n_{p}})^{\prime}\vert \frac{d\mu}{\mu(J)}\geq \frac{1}{\mu(J)} \int_{J} \lambda d\mu. (\star) \end{align} By assumption on $f$ , there exists a positive real number $M$ such that $M = \sup_{x \in J} f (x)$. Moreover, for all $p \geq 0$, $T^{n_{p}}$ is monotone on $J$ according to a Proposition in the paper. Consequently, \begin{align*} \int_{J} \vert (T^{n_{p}})^{\prime} \vert \ d\mu &=\int_{J} \vert(T^{n_{p}})^{\prime}(x) \vert f(x) dx \\ &\leq M \int_{J} \vert (T^{n_{p}})^{\prime}(x) \vert dx\\ &= M\vert \int_{J} (T^{n_{p}})^{\prime}(x) dx \vert \\ & = M \vert T^{n_{p}}(s_{2}) - T^{n_{p}}(s_{1}) \vert \quad , \quad J=(s_{1} , s_{2}) \\ &\leq M \end{align*} With $ \star $, one deduces that \begin{align*} \int_{J} \lambda d\mu \leq 0 \end{align*} and, since $μ(J )> 0$, that \begin{align*} \mu \big( \lbrace x \in [0,1] : \lambda(x) >0 \rbrace \big) < 1 \end{align*} This concludes the proof of the proposition.

My questions :
1- How can I deduce $ \int_{J} \lambda d\mu \leq 0 $ ?
2- Is the proof done by contradiction ?

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Since $\int_J |(T^{n_p})'|\,d\mu \le M$ as argued, we have $\log \int_{J} \vert (T^{n_{p}})^{\prime}\vert \frac{d\mu}{\mu(J)} \le \log \frac{M}{\mu(J)}$. Hence $(\star)$ implies $$\frac{1}{\mu(J)} \int_J \lambda\,d\mu \le \liminf_{p \to \infty} \frac{1}{n_p} \log \frac{M}{\mu(J)} = 0$$ because $n_p \to \infty$ as $p \to \infty$.

The overall structure of the proof seems complete to me, though I would call this more of a proof by contrapositive rather than by contradiction. We start by supposing it is not true that $\max_{j\in \lbrace 0 , 1 ,\cdots ,\mathit{l}_{n}\rbrace} \big(diam(I_{j}^{n})\big) \to 0$, in which case its limsup must be greater than zero (since the liminf is obviously at least zero). We finish by showing that it is not the case that $\lambda > 0$ $\mu$-a.s. (because there is a set of positive measure, namely $J$, over which its integral is non-positive).

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