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Is this function $f:\Bbb R\to \Bbb R$ given by $f(x)=\lim f_n(x)$

where $$f_n(x)= \begin{cases} \dfrac{n}{n+1}&\text{;$x\in \Bbb Q^c$}\\0 &;x\in \Bbb Q\end{cases}$$Lebesgue Integrable?

My try:

I got $$f(x)=\begin{cases} \dfrac{n}{n+1}& \text{;$x\in \Bbb Q^c$}\\0& ;x\in \Bbb Q\end{cases}$$

and hence measurable

But $\int_\Bbb R f=\int _{\Bbb Q^c} f=m(\Bbb Q^c)=\infty$.So not integrable

Is my answer correct?

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    The limit function is $f=1_\mathbb{Q^c}$, which is one almost everywhere, hence it is not integrable on any set of unbounded measure. – 2017-02-05

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The limit function $f$ is just $1$ (a.e.) (just calculate the limit pointwise) so it is trivially not integrable over $\mathbb{R}$. Without actually calculating that $f=1$ you could also argue that, in this case, $f=\sup_nf_n$ so, since the $f_n$'s are measurable, $f$ is measurable; and, for the integrability just notice that $$\int_\mathbb{R}f\,dm\geq\int_\mathbb{R}f_1\,dm=\frac{1}{1+1}m(\mathbb{R}).$$

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    @learningmaths And, no, your answer is not correct since the limit function $f$ can't depend on $n$. The limit function is $1$ for irrationals and $0$ for irrationals. – 2017-02-05