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Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a Lebesgue measurable function and $\phi(x)=\lambda ( \lbrace{ t: f(t) >x \rbrace} )$. Prove that $\phi$ is right-continuous but not necessarily left-continuous.

I was thinking this: to every measure $\mu$ I can associate a right-continuous monotone function such that if $\mu([a,b]) < \infty$ then $\mu([a,b])=F(b)-F(a)$. In this case, the Lebsesgue measure, the function would be the identity function. Is this argument related to the exercise? And if yes how can I continue?

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Let $x\in\mathbb{R}$ be fixed, and consider a sequence $(x_n)_{n\in\mathbb{N}}$ that tends to $x$ by above.

Then,

$$\left|\phi(x)-\phi(x_n)\right|=\lambda\left(\left\{t\ :\ x<f(t)\le x_n\right\}\right)=\lambda\left(f^{-1}(]x,x_n])\right).$$

Now, the sequence $(]x,x_n])_{n\in\mathbb{N}}$ is a decreasing sequence of Borel intervals, hence $(f^{-1}(]x,x_n]))_{n\in\mathbb{N}}$ is a decreasing sequence of Lebesgue-measurable sets.

Note that $\bigcap_{n\in\mathbb{N}}f^{-1}\left(]x,x_n]\right)=f^{-1}\left(\bigcap_{n\in\mathbb{N}}]x,x_n]\right)=\emptyset$, hence by continuity of $\lambda$ at $0$,

$$\lim_{n\rightarrow+\infty}\left|\phi(x)-\phi(x_n)\right|=\lim_{n\rightarrow+\infty}\lambda\left(f^{-1}(]x,x_n])\right)=\lambda\left(\bigcap_{n\in\mathbb{N}}f^{-1}(]x,x_n])\right)=0.$$

To prove that the converse does not necessarily hold, consider $f=\widetilde{x}\in\mathbb{R}$ which is constant.

Let us show that $\phi$ is not left-continuous at $\widetilde{x}$. Consider a sequence $(x_n)_{n\in\mathbb{N}}$ that tends to $\widetilde{x}$ by below.

Similarly, for any $n\in\mathbb{N}$,

$$\left|\phi(\widetilde{x})-\phi(x_n)\right|=\lambda\left(\left\{t\ :\ x_n<f(t)\le \widetilde{x}\right\}\right)=\lambda\left(\mathbb{R}\right)=+\infty,$$

which can thus not tend to zero.

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Thanks for your answer. It is really clear. –  user73793 May 2 at 8:51
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I just want to be sure about the notation...I guess that your $x_0$ it's just a costant that should be different from the first term $x_0$ in the sequence. –  user73793 May 2 at 9:48
    
Oh right, that notation is not so smart. $x_0$ is just a constant which has no link with the first term of the sequence. I'll call it $\widetilde{x}$ to make it easier to read. –  Ian May 2 at 10:10
    
Ok perfect. Thanks again. –  user73793 May 2 at 12:48

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