Showing that $\{f_n \}$ converges to $f$ is equivalent to $\lim_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0$

Question

Let $(X,\mathcal{M},\mu)$ be a measure space. We say that $\{f_n\}$ converges to $f$ in measure if, for any $\epsilon>0$ $$ \lim\limits_{n \to \infty} \mu\Big(\{x\in X: |f_n(x)-f(x)| \ge \epsilon \}\Big)=0.$$

1. Show that $\{f_n \}$ converges to $f$ in measure if $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0.$$

2. Suppose $\mu(X) < \infty$. Show that $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0$$ if $\{ f_n \}$ converges to $f$ in measure.

Answer 1

You have to assume $\mu(X)<\infty$. Without loss of generality, let $f_n\to 0$ in measure. Then for any $\varepsilon>0$, $\lim_{n\to\infty}\mu(\{x\in X: |f_n(x)|\ge \varepsilon\})=0$. Note that the function $\frac{t}{1+t}$ is increasing for $t>-1$. So $$\int_X\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)=\int_{|f_n|\ge\varepsilon}\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)+\int_{|f_n|\le\varepsilon}\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)\le\int_{|f_n|\ge\varepsilon}1d\mu(x)+\int_{|f_n|\le\varepsilon}\frac{\varepsilon}{1+\varepsilon}d\mu(x)\le \mu({|f_n|\ge\varepsilon})+\frac{\varepsilon}{1+\varepsilon}\mu(X).$$ Thus for $\forall \varepsilon>0$, $$ \overline{\lim}_{n\to\infty}\int_X\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)\le \frac{\varepsilon}{1+\varepsilon}\mu(X)$$ and hence $$ \overline{\lim}_{n\to\infty}\int_X\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)=0.$$ Conversely, note, for $\forall\varepsilon>0$, \begin{eqnarray*} \int_X\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)&=&\int_{|f_n|\ge\varepsilon}\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)+\int_{|f_n|\le\varepsilon}\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)\\ &\ge&\int_{|f_n|\ge\varepsilon}\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)\\ &\ge&\int_{|f_n|\ge\varepsilon}\frac{\varepsilon}{1+\varepsilon}d\mu(x)\\ &=&\frac{\varepsilon}{1+\varepsilon}\mu(|f_n|\ge\varepsilon). \end{eqnarray*} Thus if $\lim_{n\to\infty}\int_X\frac{|f_n(x)|}{1+|f_n(x)|}d\mu(x)=0$, then $\lim_{n\to\infty}\mu(|f_n|\ge\varepsilon)=0$ or $f_n\to 0$ in measure.