Example of composition of continuous function with sequence of functions which converges in measure need not converge in measure

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 ?

• Your proof in the uniformly continuous case is correct. – PhoemueX Apr 7 at 6:40

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.

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.