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Let $$C_{0}(\mathbb R)= \big\{\,f:\mathbb R \to \mathbb C\,\,\, \text{continuous and}\,\, \lim_{x\to \pm \infty}f(x)=0 \big\}.$$

Assume that $f\in C_{0}(\mathbb R)$.

My question is: Is it always true that, $\int_{\mathbb R}|f(x)| dx < \infty ? $ If not, counter example ?

(My attempt: Since $f\in C_{0}(\mathbb R)$, so given $\epsilon >0$ there is a compact set $K\subset \mathbb R$ and $M> 0$ such that $|f(x)|\leq M$ for every $x\in K$ and $|f(x)|< \epsilon $ for every $x\in \mathbb R - K$; thus, $\int_{\mathbb R}|f(x)| dx = \int_{K}|f(x)| dx + \int_{\mathbb R -K} |f(x)|dx \leq C + \epsilon \mu(\mathbb R - K)$, where $\mu$ is Lebesgue measure on $\mathbb R$ ; but, certainly, this is incomplete argument !!)

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    $\begingroup$ Let $f$ be some continuous function (defined piecewise) that is eventually $1/x$. $\endgroup$ Jan 5, 2014 at 17:31
  • $\begingroup$ It is not true. I'm not sure whether this is homework so I'll just say that much. Look for a counterexample. There are some very simple ones! Have fun. $\endgroup$
    – user119261
    Jan 5, 2014 at 17:49
  • $\begingroup$ False, this has also been discussed on math.stackexchange.com/questions/663230/… $\endgroup$
    – michek
    Feb 4, 2014 at 14:21

1 Answer 1

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Let for example $$ f(x)=\frac{1}{|x|+1}. $$ Then $$ \int_{-\infty}^\infty f(x)\,dx=\infty. $$

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