# A question about measure theory

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 ?

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).