Show that it is metric and the measurable is 0

[mathmari]

Hey!

In a space of finite measure, if $f$ and $g$ are measurable we set $\rho (f,g)=\int \frac{|f-g|}{1+|f-g|}d \mu$.

Show that $\rho$ is metric and that $f_n \rightarrow f$ as for $\rho$ if and only if $\forall c>0$ we have that $\mu(\{|f_n-f|>c\})\rightarrow 0$.What does "$f_n \rightarrow f$ as for $\rho$" mean ?? (Wondering)

 

[Opalg]

"$f_n \rightarrow f$ as for $\rho$" is not idiomatic English. But it must surely mean "$\rho(f_n,f) \to0$".

 

[Euge]

To prove that $\rho$ is a metric, first show that the function $D$ given by

\(\displaystyle D(f,g) = \frac{|f - g|}{1 + |f - g|}\)

is a metric.

Suppose $\rho(f_n,f) \to 0$. Fix $c > 0$. Since the function $F(x) = \frac{x}{1 + x}$ is strictly increasing on $[0, \infty)$,

\(\displaystyle \mu\{|f_n - f| > c\} \le \mu\left\{D(f_n,f) > \frac{c}{1 + c}\right\}.\)

By Chebyshev's inequality, it follows that

\(\displaystyle \mu\left\{D(f_n,f) > \frac{c}{1 + c}\right\} \le \frac{1 + c}{c} \int D(f_n, f)\, d\mu = \frac{1+c}{c}\rho(f_n,f) \to 0.\)

Therefore $\mu\{|f_n - f| > c\} \to 0$, and consequently $f_n$ converges in $\mu$-measure to $f$.

To prove the converse, I'll let $X$ be the common domain of $f$ and the $f_n$. This is where the hypothesis $\mu(X) < \infty$ is essential. Suppose $f_n$ converges in $\mu$-measure to $f$. Given $\varepsilon > 0$, set $E_n(\varepsilon) = \{|f_n - f| > \varepsilon\}$, $n = 1, 2, 3,\ldots.$ Then

\(\displaystyle \rho(f_n,f) = \int_{E_n(\varepsilon)} D(f_n,f)\, d\mu + \int_{X\setminus E_n(\varepsilon)} D(f_n,f)\, d\mu.\)

Since $D(f_n,f) < 1$,

\(\displaystyle \int_{E_n(\varepsilon)} D(f_n,f)\, d\mu < \mu(E_n(\varepsilon)).\)

On $X\setminus E_n(\varepsilon)$, $D(f_n,f) \le \frac{\varepsilon}{1 + \varepsilon}$. Thus

\(\displaystyle \int_{X\setminus E_n(\varepsilon)} D(f_n,f)\, d\mu \le \frac{\varepsilon}{1 + \varepsilon}\mu(X\setminus E_n(\varepsilon)) \le \frac{\varepsilon}{1 + \varepsilon}\mu(X). \)

Hence

\(\displaystyle \limsup_{n\to \infty}\rho(f_n,f) \le \limsup_{n\to \infty} \left(\mu(E_n(\varepsilon)) + \frac{\varepsilon}{1 + \varepsilon}\mu(X)\right) \le \frac{\varepsilon}{1 + \varepsilon}\mu(X),\)

since $\mu(E_n(\varepsilon)) \to 0$ by hypothesis. As $\mu(X) < \infty$ and $\varepsilon$ was arbitrary positive number, $\rho(f_n,f) \to 0$.

 

[mathmari]

To prove that $\rho$ is a metric, first show that the function $D$ given by

\(\displaystyle D(f,g) = \frac{|f - g|}{1 + |f - g|}\)

is a metric.

To show that $D(f,g)$ is metric I have done the following:

1. $$D(f,g)\geq 0, \forall f,g$$
   
2. $$D(f,g)=0\Leftrightarrow |f-g|=0\Leftrightarrow f=g$$
   
3. $$D(g,f)=\frac{|g-f|}{1+|g-f|}=\frac{|f-g|}{1+|f-g|}=D(f,g)$$
   
4. $$D(f,g)=\frac{|f-g|}{1+|f-g|}=\frac{|f-z+z-g|}{1+|f-g|}\leq \frac{|f-z|+|z-g|}{1+|f-g|}=\frac{|f-z|}{1+|f-g|}+\frac{|z-g|}{1+|f-g|}$$ How can I continue to show the triangle inequality?? (Wondering)

 

[Euge]

To show that $D(f,g)$ is metric I have done the following:

1. $$D(f,g)\geq 0, \forall f,g$$
   
2. $$D(f,g)=0\Leftrightarrow |f-g|=0\Leftrightarrow f=g$$
   
3. $$D(g,f)=\frac{|g-f|}{1+|g-f|}=\frac{|f-g|}{1+|f-g|}=D(f,g)$$
   
4. $$D(f,g)=\frac{|f-g|}{1+|f-g|}=\frac{|f-z+z-g|}{1+|f-g|}\leq \frac{|f-z|+|z-g|}{1+|f-g|}=\frac{|f-z|}{1+|f-g|}+\frac{|z-g|}{1+|f-g|}$$ How can I continue to show the triangle inequality?? (Wondering)

Hi mathmari,

To prove the triangle inequality, consider the function $\phi(x) := \frac{x}{1 + x}$ on $[0, \infty)$. Show that $\phi$ is increasing and satisfies the inequality $\phi(x + y) \le \phi(x) + \phi(y)$ for all $x, y \in [0,\infty)$. Then you may establish the inequalities

\(\displaystyle D(f,g) = \phi(|f - g|) \le \phi(|f - z| + |z - g|) \le \phi(|f - z|) + \phi(|z - g|) = D(f,z) + D(z,g)\)

for all $f, g, z$.

 

[mathmari]

Hi mathmari,

To prove the triangle inequality, consider the function $\phi(x) := \frac{x}{1 + x}$ on $[0, \infty)$. Show that $\phi$ is increasing and satisfies the inequality $\phi(x + y) \le \phi(x) + \phi(y)$ for all $x, y \in [0,\infty)$. Then you may establish the inequalities

\(\displaystyle D(f,g) = \phi(|f - g|) \le \phi(|f - z| + |z - g|) \le \phi(|f - z|) + \phi(|z - g|) = D(f,z) + D(z,g)\)

for all $f, g, z$.

Ok! (Smile)

Having shown that $D$ is metric, how do we conclude that $\rho$ is metric?? (Wondering)

 

[Euge]

Ok! (Smile)

Having shown that $D$ is metric, how do we conclude that $\rho$ is metric?? (Wondering)

The only property you really need from $D$ is the triangle inequality, so that you can prove the triangle inequality for $\rho$. Symmetry and positive definiteness of $\rho$ follow directly from the definition of $\rho$ and integral properties. Moreover, since $\rho(f,g)$ is an integral over a space of finite measure and $f$ and $g$ are measurable, $\rho(f,g)$ is finite.

 

[mathmari]

Suppose $\rho(f_n,f) \to 0$. Fix $c > 0$. Since the function $F(x) = \frac{x}{1 + x}$ is strictly increasing on $[0, \infty)$,

\(\displaystyle \mu\{|f_n - f| > c\} \le \mu\left\{D(f_n,f) > \frac{c}{1 + c}\right\}.\)

By Chebyshev's inequality, it follows that

\(\displaystyle \mu\left\{D(f_n,f) > \frac{c}{1 + c}\right\} \le \frac{1 + c}{c} \int D(f_n, f)\, d\mu = \frac{1+c}{c}\rho(f_n,f) \to 0.\)

Therefore $\mu\{|f_n - f| > c\} \to 0$, and consequently $f_n$ converges in $\mu$-measure to $f$.

Could you explain it further to me?? (Wondering)

 

[Euge]

Hi mathmari,

Recall that a positive measure $\mu$ satisfies subadditivity: If $A$ and $B$ are measurable and $A \subset B$, then $\mu(A) \le \mu(B)$. Since $F$ is strictly increasing on $[0,\infty)$, $|f_n - f| > c$ implies $F(|f_n - f|) > F(c)$, i.e., $D(f_n, f) > \frac{c}{1 + c}$. Hence $\{x \in X : |f_n(x) - f(x)| > c\} \subset \{x\in X : D(f_n(x),f(x)) > \frac{c}{1 + c}\}$. Consequently by subadditivity, $\mu\{|f_n - f| > c\} \le \mu\{D(f_n,f) > \frac{c}{1 + c}\}$.

For the next part, Chebyshev's inequality was used -- for any $g \in L^1(X,\mu)$, $\mu\{|g| > \alpha\} \le \alpha^{-1} \int_X |g|\, d\mu$. Now set $g = D(f_n,f)$ and $\alpha = \frac{c}{1 + c}$ to get

\(\displaystyle \mu\left\{D(f_n,f) > \frac{c}{1 + c}\right\} \le \frac{1 + c}{c} \int_X D(f_n,f)\, d\mu = \frac{1 + c}{c}\rho(f_n,f).\)

 

[mathmari]

Since $F$ is strictly increasing on $[0,\infty)$, $|f_n - f| > c$ implies $F(|f_n - f|) > F(c)$, i.e., $D(f_n, f) > \frac{c}{1 + c}$.

Could you explain me further this part?? (Wondering)

 

[Euge]

Sorry, I may have confused you with the notation. Since $F$ is strictly increasing on $[0,\infty)$, $F(x) > F(y)$ whenever $x > y \ge 0$. So for each $x \in X$, $|f_n(x) - f(x)| > c$ implies $F(|f_n(x) - f(x)|) > F(c)$. By definition of $D$, $D(f_n(x), f(x)) = F(|f_n(x) - f(x)|)$. Also, $F(c) = \frac{c}{1 + c}$. So we have that for all $x \in X$, $|f_n(x) - f(x)| > c$ implies $D(f_n(x),f(x)) < \frac{c}{1 + c}$. In terms of sets, this translates to

\(\displaystyle \{x\in X : |f_n(x) - f(x)| > c\} \subset \left\{x \in X : D(f_n(x), f(x)) > \frac{c}{1 + c}\right\}.\)

 

[mathmari]

Sorry, I may have confused you with the notation. Since $F$ is strictly increasing on $[0,\infty)$, $F(x) > F(y)$ whenever $x > y \ge 0$. So for each $x \in X$, $|f_n(x) - f(x)| > c$ implies $F(|f_n(x) - f(x)|) > F(c)$. By definition of $D$, $D(f_n(x), f(x)) = F(|f_n(x) - f(x)|)$. Also, $F(c) = \frac{c}{1 + c}$. So we have that for all $x \in X$, $|f_n(x) - f(x)| > c$ implies $D(f_n(x),f(x)) < \frac{c}{1 + c}$. In terms of sets, this translates to

\(\displaystyle \{x\in X : |f_n(x) - f(x)| > c\} \subset \left\{x \in X : D(f_n(x), f(x)) > \frac{c}{1 + c}\right\}.\)

Could you explain me the last inclusion?? (Wondering)

Why does it stand that $$\{x\in X : |f_n(x) - f(x)| > c\} \subset \left\{x \in X : D(f_n(x), f(x)) > \frac{c}{1 + c}\right\}$$ (Wondering)