Step in proof of Birkhoff ergodic theorem

Question

I have a question about the proof of Birkhoff ergodic theorem in the book : An Introduction to Ergodic theory by Peter Walters.

At page 38 he tells us that :

For reals $\alpha, \beta$ we define $E_{\alpha, \beta} := \{ x \in X \mid f_*(x) < \beta ~\text{and}~ f^*(x) > \alpha\}$.

[...]

From corollary 1.16.1 we get $$ \int_{E_{\alpha, \beta}} f dm = \int_{E_{\alpha, \beta} \cap B_\alpha} f dm \geq \alpha m(E_{\alpha, \beta} \cap B_\alpha) = \alpha m(E_{\alpha, \beta}). $$ If we replace $f, \alpha, \beta$ by $-f, -\beta, -\alpha$, respectively, then since $(-f)^\ast = -f_\ast$ and $(-f)^\ast = -f_\ast$ we get $$ \int_{E_{\alpha, \beta}} f dm \leq \beta m(E_{\alpha, \beta}). $$

But basically if we do the substitution we get $$ \int_{E_{-\beta, -\alpha}} f dm \leq \beta m(E_{-\beta, -\alpha}). $$ So my question is : how we can go from $E_{-\beta, -\alpha}$ to $E_{\alpha, \beta}$ ?

Answer 1

Note that $$ E_{-\beta,-\alpha}(-f) := \left\{ x \in X \mid (-f)_*(x) < -\alpha ~\text{and}~ (-f)^*(x) > -\beta\right\}\\ =\left\{ x \in X \mid -(f^*)(x) < -\alpha ~\text{and}~ -(f)_*(x) > -\beta\right\}\\ =\left\{ x \in X \mid (f^*)(x) >\alpha ~\text{and}~ (f)_*(x) <\beta\right\} =E_{\alpha,\beta}(f), $$ where the second equality follows from your observations that $(-f)_*=-(f^*)$ and $(-f)^*=-(f_*)$.