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Example $\phi $ is continuous function $f_n \to f$ in measure. with $\phi \circ f_n\not\to \phi \circ f$ in measure

I am trying to find above example but unable to construct.

Also I tried following theorem

$\phi $ is uniformly continuous function $f_n \to f$ in measure. with $\phi \circ f_n\to \phi \circ f$ in measure

My attempt:

Let $\phi $ is uniformly continuous

$\forall \epsilon >0$ $\exists \delta >0$ such that $|x-y|<\delta\implies |\phi (x)-\phi (y)|<\epsilon $

$E=\{x||f_n(x)-f(x)|>\delta \}$

$\mu(E)\to 0$

On $E^c$ $|f_n(x)-f(x)|\leq \delta $$\implies |\phi \circ f_n(x)-\phi \circ f(x)||<\epsilon $

SO it is converges in measure .

Is there is any mistake ?

Please Help me to construct example

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    $\begingroup$ Your proof in the uniformly continuous case is correct. $\endgroup$ – PhoemueX Apr 7 at 6:40
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The answer by @Kavi Rama Murthy is valid on finite measure spaces.

On a general measure space, you can find a counterexample. Take for example the real line with the Lebesgue measure, and $f_n (x) = x + 1/n$, $f(x) = x$, and $\phi (x) = x^2$.

Then $\phi(f_n(x))-\phi(f(x)) = 2x/n +n^{-2}$, which has absolute value at least one on the set $[n,\infty)$. Directly from the definition, this shows that you don't have $\phi \circ f_n \to \phi \circ f$ in measure.

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Your proof for the case when $\phi$ is uniformly continuous is fine.

There is no counterexample when uniform continuity is replaced by continuity. If $f_n \to f$ in measure and if $\phi$ is continuous then $\phi\circ f_n \to \phi \circ f$ in measure. This follows from two facts:

1) A sequence of real numbers tends to $0$ iff every subsequence of it has a further subsequence which tends to $0$.

2) Convegergence in measure implies almost everywhere convergence for a subsequence.

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