6
$\begingroup$

Let $(X,\mathcal{M},\mu)$ be a measurable space, suppose $\mu(X)<\infty$. If $f$ and $g$ measurable functions on $X$, define $$\rho(f,g):=\int \frac{|f-g|}{1+|f-g|}d\mu.$$

Let $(f_{n})_{n\in\mathbb{N}}$ be a sequence of measurable function. Show that $\rho(f_{n},f)\rightarrow 0$ iff $f_{n}\rightarrow f$ in measure.

My problem: Show that $f_{n}\rightarrow f$ in measure $\Longrightarrow$ $\rho(f_{n},f)\rightarrow 0$.

My attempt: Since $f_{n}\rightarrow f$ in measure, then, in particular, for each $\varepsilon>0$ we have $$\lim_{n\rightarrow \infty}\mu\left(\left\{x\in X \: \left|\: |f_{n}(x)-f(x)|\geq\frac{\varepsilon}{2\mu(X)}\right. \right\}\right). $$

Therefore, there exists $N$ such that if $n\geq N$ then $$\mu\left(\left\{x\in X \: \left|\: |f_{n}(x)-f(x)|\geq\frac{\varepsilon}{2\mu(X)}\right. \right\}\right)<\frac{\varepsilon}{2}. \tag{1}$$ We called $$E_{\varepsilon}^{n}:=\left\{x\in X \: \left|\: |f_{n}(x)-f(x)|\geq \frac{\varepsilon}{2\mu(X)}\right. \right\}.$$ Then we have $$\mu(E_{\varepsilon}^{n})<\frac{\varepsilon}{2}. \tag{2}$$ Note that $$\frac{|f-g|}{1+|f-g|}< \frac{\varepsilon}{2\mu(X)} \qquad \mbox{in }X\setminus E_{\varepsilon}^{n}. $$

So, as $\frac{|fn-f|}{1+|f_{n}-f|}\leq 1$ we have $$ \begin{array}{rcl} \rho(f_{n},f) &=& \int \frac{|f_{n}-f|}{1+|f_{n}-f|}d\mu \\ &=&{\displaystyle \int_{X\setminus E_{\varepsilon}^{n}} \frac{|f_{n}-f|}{1+|f_{n}-f|}d\mu+\int_{E_{\varepsilon}^{n}} \frac{|f_{n}-f|}{1+|f_{n}-f|}d\mu } \\ &\leq& {\displaystyle \frac{\varepsilon}{2\mu(X)}\int_{X\setminus E_{\varepsilon}^{n}}d\mu + \int_{ E_{\varepsilon}^{n}} d\mu } \\ &=& {\displaystyle \frac{\varepsilon}{2\mu(X)} \mu(X\setminus E_{\varepsilon}^{n}) +\mu( E_{\varepsilon}^{n}) } \\ &\leq& {\displaystyle \frac{\varepsilon}{2\mu(X)} \mu(X) +\mu( E_{\varepsilon}^{n}) } \\ &<& {\displaystyle \frac{\varepsilon}{2} + \frac{\varepsilon}{2} } \\ &=& \varepsilon. \end{array} $$

My Question: Is there any error in my attempt?

The doubt is born because in this link (see pag 5) makes an impractical proof and this is the only demonstration I found on the internet.

$\endgroup$
  • 1
    $\begingroup$ Your proof seems good. To be nitpicking, your $\rho$ should be $$ \rho(f, g) = \int \frac{|f-g|}{1+|f-g|} \, d\mu $$ $\endgroup$ – Sangchul Lee May 19 '17 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.