$F(x):=\int_{[a,x)} f(t) \lambda(dt)$ is absolutely continuous for all $f \in L^1(\lambda)$.

Question

I am looking at the solution to the following exercise from Rene Schilling's Measures, Integrals and Martingales. However, below in the second inequality for $|F(y_n)-F(x_n)|$, I cannot figure out why $\int_{[x_n,y_n)\cap \{|f|>R\}}|f(t)|\lambda(dt) \le \lambda(\{|f|>R\}\cap [x_n,y_n))$. Here we don't have $|f|\le 1$, so I don't think this actually works. But the rest of the proof relies on this inequality. How can I resolve this problem?

$F(x):=\int_{[a,x)} f(t) \lambda(dt)$ is absolutely continuous for all $f \in L^1(\lambda)$.

Answer 1

We know that $f\in L^{1}(\lambda)$ implies that $\displaystyle\int|f|d\lambda-\int_{|f|\leq R}|f|d\lambda\rightarrow 0$ as $R\rightarrow\infty$, this is just Monotone Convergence Theorem.

Therefore, we have $\displaystyle\int_{|f|>R}|f|d\lambda\rightarrow 0$ as $R\rightarrow\infty$. Pick an $R>0$ large enough that $\displaystyle\int_{[x_{n},y_{n})\cap\{|f|>R\}}|f|d\lambda\leq\int_{|f|>R}|f|d\lambda$ small enough.