Let $(X,S,\mu)$ is a measure space and $\mu(X)<\infty$. Define $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ is a metric on the space of measurable functions

Question

Let $(X,S,\mu)$ is a measure space and $\mu(X)<\infty$. Define $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ is a metric on the space of measurable functions.

My work-

symmetry,

let's show $d(f,g)=d(g,f)$

$d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ and $d(g,f)=\int\frac{|g-f|}{1+|g-f|}d\mu$

It's obvious that, $d(f,g)=d(g,f)$

My concern is triangular inequality,

What want to show is, $d(f,h)\leq d(f,g)+d(g,h)$

so the right hand side becomes,

$\int\frac{|f-g|}{1+|f-g|}d\mu+\int\frac{|g-h|}{1+|g-h|}d\mu= \int\frac{|f-g|}{1+|f-g|}+\frac{|g-h|}{1+|g-h|}d\mu$

I don't think simplification of right side will give the result easily. Can someone give me a hints. Thank you in advance

Answer 1

Let $(X,S,\mu)$ is a measure space and $\mu(X)<\infty$. Define $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$. Then $d$ is a metric on the space of measurable functions.

Let us prove a more general result.

Let $d: \Bbb R \times \Bbb R \rightarrow [0, +\infty) $ be any metric in $\Bbb R$.

Let us define $d_0: \Bbb R \times \Bbb R \rightarrow [0, 1) $ by $$d_0(x,y) = \frac{d(x,y)}{1+d(x,y)}$$ Then $d_0$ is also a metric in $\Bbb R$. In fact

1. $d_0(x,y) = 0$ iff $d(x,y) = 0$ iff $x=y$.
2. $d_0(y,x) = \frac{d(y,x)}{1+d(y,x)}= \frac{d(x,y)}{1+d(x,y)} = d_0(x,y)$
3. For triangular inequality, let $f: [0, +\infty) \rightarrow \Bbb [0,1)$ defined as $f(t) = \frac{t}{1+t}$. Note that $f'(t) = \frac{1}{(1+t)^2}>0$ so $f$ is strictly increasing. Note that $d_0= f \circ d$. So we have \begin{align*} d_0(x,z)&= f(d(x,z))\leq f(d(x,y) +d(y,z)) = \frac{d(x,y) +d(y,z)}{1+d(x,y) +d(y,z)} = \\ & = \frac{d(x,y) }{1+d(x,y) +d(y,z)}+ \frac{d(y,z)}{1+d(x,y) +d(y,z)} \leq \\ & \leq \frac{d(x,y) }{1+d(x,y)}+ \frac{d(y,z)}{1+d(y,z)} = d_0(x,y) + d_0(y,z) \end{align*} So, $d_0$ is a metric in $\Bbb R$ (actually $d_0$ is equivalent to $d$, but we don't need to prove this for this question).

Now, let $\mathcal{M}(X,S)$ be the set of measurable functions defined in $(X,S)$. Let $d_\mu: \mathcal{M}(X,S) \times \mathcal{M}(X,S) \rightarrow [0, \infty)$ be defined by $$ d_\mu (f,g) =\int d_0(f(x),g(x)) d\mu(x) = \int \frac{d(f(x),g(x))}{1+d(f(x),g(x))}d\mu(x)$$

Note that, since, for all $x$, $d_0(f(x),g(x))\in [0,1)$ and $\mu(X) <+\infty$, we have that, for all $f, g \in \mathcal{M}(X,S)$, $d_\mu (f,g) \leq \mu(X) <+\infty$. So, $d_\mu (f,g) \in [0, +\infty)$.

In order that $d_\mu$ be a metric, we must make the convention that, if $f,g \in \mathcal{M}(X,S)$ and $f=g$ a.e., we are going to consider that $f=g$ (technically we are taking the quotient of $\mathcal{M}(X,S)$ by the equivalence relation $=$ a.e.).

Let us prove that $ d_\mu$ is a metric.

1. $d_\mu (f,g) = 0$ iff $d_0(f(x),g(x))=0$ a.e iff $f=g$ a.e.
2. $ d_\mu (g,f) =\int d_0(g(x),f(x)) d\mu(x) = \int d_0(f(x),g(x)) d\mu(x)= d_\mu (f,g) $

\begin{align*} d_\mu (f,h) &= \int d_0(f(x),h(x)) d\mu(x) \leq \\ & \leq \int d_0(f(x),g(x)) d\mu(x) + \int d_0(g(x),h(x)) d\mu(x) =\\ & = d_\mu (f,g)+ d_\mu (g,h) \end{align*} So, $d_\mu$ is a metric in $\mathcal{M}(X,S)$.

The result in the question is just the special case where $d(x,y)=|x-y|$.

Remark: Note that in addition to the triangular inequality for $d_\mu$ (which depends on the triangular inequality for $d_0$), there are two other subtle points:

1. We use the fact that $\mu(X)<\infty$ to ensure that $d_\mu$ is finite. Having $d_\mu$ finite is a condition for $d_\mu$, as defined in the question, to be a metric.

2. If $f,g \in \mathcal{M}(X,S)$ and $f=g$ a.e., we must identify $f$ and $g$ (technically we are taking the quotient of $\mathcal{M}(X,S)$ by the equivalence relation $=$ a.e.). This is needed in order that $d_\mu$, as defined in the question, is a metric (otherwise $d_\mu$ would be only a pseudo-metric).

Answer 2

The only thing missing is to check that $$ \rho(x,y)=\frac{|x-y|}{1+|x-y|} $$ is a metric on $\mathbb{R}$. Here is a simple proof:

Consider the function $$ f(t)=\frac{t}{1+t}, \qquad t\geq0$$ notice that $f(t)=0$ iff $t=0$, $f$ is monotone non decreasing on $[0,\infty)$, and that $\rho(x,y)=f(|x-y|)$. That $\rho$ satisfies the triangle inequality is a consequence of $$ f(t+s)\leq f(t)+f(s),\qquad t,s\geq0$$ which follows from \begin{align} f(t+s)&=\frac{s+t}{1+s+t}=\frac{t}{1+t}\frac{1+t}{1+t+s} +\frac{s}{1+s}\frac{1+s}{1+t+s}\\ &\leq \frac{t}{1+t}+\frac{s}{1+s} \end{align}

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Though the is not needed for the conclusions in your problem, it is worthwhile noticing that on any matrix space $(S,d)$, any continuous monotone nondecreasing function $\phi:[0,\infty)\rightarrow[0,\infty)$ such that

1. $\phi(t)=0$ iff $t=0$,
2. $\phi(t+s)\leq \phi(t)+\phi(s)$ (subadditve)

induces another metric on $S$ $$\rho_\phi(x,y)=\phi(d(x,y))$$ that is equivalent to $d$.

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Answer 3

As $\mu$ is not mentioned to be signed, I am assuming it is positive measure i.e. $\mu : X \rightarrow [0,\infty)$.

Here is an approach to the Triangle Inequality :

$\textit{Case I}$ : $f=g=h$

Then $$d(f,h) = d(f,g)=d(g,h)=0$$ $$\implies d(f,h)=d(f,g)+d(g,h)$$

$\textit{Case II}$ : $f=g\neq h$ or $f \neq g=h$

W.L.O.G. assume $f=g \neq h$

Then $$d(f,g)=0$$ $$d(f,h)=d(g,h)$$ $$\implies d(f,h)=d(f,g)+d(g,h)$$

$\textit{Case III}$ : $f\neq g\neq h$

Then $$|f-g|, |g-h|, |f-h| > 0$$ $$|f-h| \leq |f-g|+|g-h| $$ $$|f-h|(|f-g|+|g-h|) = |f-h|(|f-g|+|g-h|) $$

Adding above two equations :

$$|f-h|(|f-g|+|g-h|) + |f-h| \leq |f-h|(|f-g|+|g-h|) + |f-g|+|g-h|$$ $$\implies |f-h|(1+|f-g|+|g-h|)\leq (1+|f-h|)(|f-g|+|g-h|)\\$$

$$\begin{align} \implies \frac{|f-h|}{1+|f-h|} &\leq \frac{|f-g|+|g-h|}{1+|f-g|+|g-h|}\\ &= \left(\frac{|f-g|}{1+|f-g|}\right) \left(\frac{1+|f-g|}{1+|f-g|+|g-h|}\right) + \left(\frac{|g-h|}{1+|g-h|}\right)\left(\frac{1+|g-h|}{1+|f-g|+|g-h|}\right)\\ &\leq \left(\frac{|f-g|}{1+|f-g|}\right) \left(\frac{1+|f-g|}{1+|f-g|}\right) + \left(\frac{|g-h|}{1+|g-h|}\right)\left(\frac{1+|g-h|}{1+|g-h|}\right)\\ &= \frac{|f-g|}{1+|f-g|} + \frac{|g-h|}{1+|g-h|}\\ \\ \end{align}$$

$$\therefore \frac{|f-h|}{1+|f-h|} \leq \frac{|f-g|}{1+|f-g|} + \frac{|g-h|}{1+|g-h|}$$

$$\implies \int \frac{|f-h|}{1+|f-h|} d\mu \leq \int\frac{|f-g|}{1+|f-g|} d\mu + \int\frac{|g-h|}{1+|g-h|} d\mu \\$$

$$\therefore d(f,h) \leq d(f,g)+d(g,h)$$