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Given a measure space $\left(X,\mathcal{F},\mu\right)$ and two $\mathcal{F}-\mbox{measurable}$ functions $f,g:\left(X,\mathcal{F}\right)\to\mathbb{R}$ we define the following:$$d\left(f,g\right)=\inf_{a>0}\left(a+\mu\left(\left\{ x\in X\;|\;\left|f\left(x\right)-g\left(x\right)\right|>a\right\} \right)\right)$$ It can be shown that $d$ is symmetric and admits the triangle inequality. I'm trying to show that given two functions $f,g:\left(X,\mathcal{F}\right)\to\mathbb{R}$ and $\alpha\in\mathbb{R}$ : $$d\left(\alpha f,\alpha g\right)\leq\max\left\{ 1,\left|\alpha\right|\right\} \cdot d\left(f,g\right)$$ I've tried going at it from a couple of directions but I always get stuck with an inequality involving $\mu\left(\left\{ x\in X\;|\;\max\left\{ 1,\left|\alpha\right|\right\} \left|f\left(x\right)-g\left(x\right)\right|>a\right\} \right)$ and no way to extract that pesky maximum outside.

Also regardless of this question I think that convergence relative to $d$ is equivalent to convergence in measure relative to $\mu$ is that true?

Help would be appreciated.

Serpahimz
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  • First take the case $\alpha = 0$ out of the way; that is very easy. Then take a sharp look to see that $d(-\alpha f,-\alpha g) = d(\alpha f,\alpha g)$, so it suffices to consider $\alpha > 0$. Then consider the cases $0 < \alpha \leqslant 1$ and $1 < \alpha$ separately. – Daniel Fischer Nov 10 '13 at 21:12
  • @DanielFischer, Thanks for the reply, I'll try what you suggested in a moment. Regarding my other question, I've been trying to prove that there is indeed such an equivalence but without success, is it even true? (if it is I'd really appreciate it if you could help me with the proof). – Serpahimz Nov 10 '13 at 22:46

2 Answers2

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Let us answer

Also regardless of this question I think that convergence relative to $d$ is equivalent to convergence in measure relative to $\mu$ is that true?

first. Yes, that is true.

Let $f_n \to f$ in measure. Let $\varepsilon > 0$ be arbitrary. By the convergence in measure, there is an $N$ such that $\mu(\{ x: \lvert f(x) - f_n(x)\rvert > \varepsilon/2\}) < \varepsilon/2$ for all $n \geqslant N$. But then we have

$$d(f,f_n) \leqslant \frac{\varepsilon}{2} + \mu\left(\left\lbrace x : \lvert f(x) - f_n(x)\rvert > \frac{\varepsilon}{2}\right\rbrace\right) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

for $n \geqslant N$. So convergence in measure implies $d$-convergence. Conversely, let $d(f,f_n)\to 0$, and let $\varepsilon > 0$ be arbitrary. We need to show that $\mu(\{ x : \lvert f(x) - f_n(x)\rvert \geqslant \varepsilon\}) \to 0$. Let $\delta > 0$ be arbitrary, subject to the restriction $\delta < \varepsilon$. Since $d(f,f_n) \to 0$, there is an $N$ with $d(f,f_n) < \delta/2$ for $n \geqslant N$. By the definition of $\inf$, there is then for each $n \geqslant N$ an $a_n > 0$ with

$$a_n + \mu(\{ x : \lvert f(x) - f_n(x)\rvert > a_n\}) < \delta.$$

But then we have $a_n < \delta$, and $\mu(\{ x : \lvert f(x) - f_n(x)\rvert > a_n\}) < \delta$, and since $a_n < \delta < \varepsilon$, that implies $\mu(\{ x : \lvert f(x) - f_n(x)\rvert \geqslant \varepsilon\}) < \delta$. That holds for all $n \geqslant N$, hence $f_n \to f$ in measure.

Regarding the inequality

$$d(\alpha f,\alpha g) \leqslant \max \{1,\lvert\alpha\rvert\}\cdot d(f,g),$$

the case $\alpha = 0$ is clear ($d(0,0) = 0$), and since multiplication by $-1$ leaves the absolute modulus unchanged, the case for negative $\alpha$ follows from that for positive $\alpha$.

Let first $0 < \alpha \leqslant 1$. Then

$$\begin{align} d(\alpha f,\alpha g) &= \inf_{a > 0} \left(a + \mu(\{ x : \lvert \alpha f(x) - \alpha g(x)\rvert > a\}) \right)\\ &= \inf_{a>0} \left(a + \mu(\{ x : \lvert f(x) - g(x)\rvert > a/\alpha\right)\\ &= \inf_{b > 0} \left(\alpha b + \mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\ &\leqslant \inf_{b > 0} \left(b + \mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\ &= d(f,g). \end{align}$$

For $\alpha > 1$, the computation is almost the same, only in the end we don't eliminate $\alpha$ and instead multiply the measure with $\alpha$,

$$\begin{align} d(\alpha f, \alpha g) &= \dotsb\\ &= \inf_{b > 0} \left(\alpha b + \mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\ &\leqslant \inf_{b > 0} \left(\alpha b + \alpha\mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\ &= \alpha d(f,g). \end{align}$$

Daniel Fischer
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  • Hey Daniel, thanks for the reply. I'm trying to use the equivalence between convergence and measure and convergence in $d$ to show that if $f_{n}$ converges to $f$ in measure and $a_{n}$ is a sequence of real numbers converging to $a\in\mathbb{R}$ then $a_{n}\cdot f_{n}$ converges to $a\cdot f$ in measure. I got to a point where for large enough $n$ I know that $d\left(a_{n},f_{n},af\right)\leq\frac{\varepsilon}{2}+d\left(a_{n}f,af\right)$. I'm having trouble showing that $d\left(a_{n}f,af\right)$ can be arbitrarily small. I'd appreciate it if you could help – Serpahimz Nov 11 '13 at 20:32
  • If $f$ is integrable, you have convergence. Since $d(\alpha_n f,\alpha f) = \inf \left(b + \mu({ \lvert \alpha_n-\alpha\rvert\cdot \lvert f(x)\rvert > b}\right)$, if $\lvert\alpha_n-\alpha\rvert \leqslant\varepsilon$, you have $\mu(\dotsc) \leqslant \varepsilon\lVert f\rVert_1/b$, and $b=\sqrt{\varepsilon}$ gives you $d(\alpha_n f,\alpha f) \leqslant \sqrt{\varepsilon}(1+\lVert f\rVert_1)$. If $f$ isn't integrable, $\alpha_n f$ need not converge to $\alpha f$ in measure. (Integrability is a stronger condition than required, weak $L^1$ suffices; but you need something.) – Daniel Fischer Nov 11 '13 at 21:07
  • I forgot to mention that I know that $f_{n}$ and $f$ are measurable, does that suffice? I haven't reached the part about Integrability through measures or $L^{p}$ spaces yet. – Serpahimz Nov 11 '13 at 21:39
  • No, unfortunately that doesn't suffice. You need that $f$ isn't large on large sets. If you consider $X=\mathbb{R}$ and $f(x) = x$, then $a_n f \to f$ in measure if and only if all but finitely many $a_n = 1$. For every $c\neq 0$ and $\varepsilon < \infty$, you have $\mu({x:\lvert c f(x)\rvert > \varepsilon}) = \infty$. You need some condition that prevents that. The existence of a $C$ such that for all $K > 0$ you have $\mu({x: \lvert f(x)\rvert > K}) \leqslant \frac{C}{K}$ is such a condition (but weaker conditions do suffice too). – Daniel Fischer Nov 11 '13 at 21:51
  • The fact $a_{n}f$ doesn't converge to $af$ doesn't necessarily mean that $a_{n}f_{n}$ doesn't converge to $af$ though. Is the original claim I was trying to prove also false? – Serpahimz Nov 12 '13 at 06:04
  • Never mind, I managed to prove it :) – Serpahimz Nov 12 '13 at 07:32
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Right I took another crack at it and I think I might have proven the claim in my second question. I'm not confident about the proof so I'd appreciate it if someone could review it.

From the definition of $d\left(f_{n},f\right)$ we know that for all $\varepsilon>0$ : $$d\left(f_{n},f\right)\leq\varepsilon+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>\varepsilon\right\} \right)$$ Since $f_{n}\to f$ in measure we know that for all $\varepsilon>0$ there is an $N$ such that for all $n>N$ :$$\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>\varepsilon\right\} \right)<\varepsilon$$ So for all $n>N$ we can see that $d\left(f_{n},f\right)<2\varepsilon$ and thus $d\left(f_{n},f\right)\overset{n\to\infty}{\longrightarrow}0$ .

On the other hand given $\varepsilon>0$ from the definition of $d\left(f_{n},f\right)$ for each $n\in\mathbb{N}$ there is an $a\left(n\right)>0$ such that:$$a\left(n\right)+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)\leq d\left(f_{n},f\right)+\frac{\varepsilon}{2}$$ Since $d\left(f_{n},f\right)\overset{n\to\infty}{\longrightarrow}0$ there is an $N$ such that $d\left(f_{n},f\right)<\frac{\varepsilon}{2}$ for all $n>N$ and thus for all $n>N$:$$a\left(n\right)+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)\leq d\left(f_{n},f\right)+\frac{\varepsilon}{2}<\varepsilon $$ Since this is true for all $\varepsilon>0$ we get that:$$\lim_{n\to\infty}\left(a\left(n\right)+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)\right)=0$$ Since this is a sum of two non-negative sequences we know that $a\left(n\right)\overset{n\to\infty}{\longrightarrow}0$ and: $$\left(\star\right)\;\lim_{n\to\infty}\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)=0$$ Now given $a>0$ there is an $N$ such that $a\left(N\right)\leq a$ and thus:$$\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{N}\left(x\right)-f\left(x\right)\right|>a\right\} \right)\leq\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{N}\left(x\right)-f\left(x\right)\right|>a\left(N\right)\right\} \right)$$ Also for all $\varepsilon>0$ from $\left(\star\right)$ there is an $N^{'}\geq N$ such that for all $n>N^{'}$ we get:$$\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\right\} \right)\leq\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(N\right)\right\} \right)<\varepsilon$$ Since this is true for all $\varepsilon>0$ we get that for all $a>0$ :$$\lim_{n\to\infty}\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\right\} \right)=0$$ And thus $f_{n}\to f$ in measure.

Serpahimz
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