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Let $(X,d)= (C[0,1],d)$ where $C[0,1]$ is the set of real-valued continuous functions on $[0,1]$ and $d= \int_{0}^{1} |f-g|$ is the Riemann Integral.

Suppose $(f_n)$ is a Cauchy sequence in $(X,d) $ , show that $ \lim_{n\to\infty} \int_{0}^{1} f_n $ exists.

My attempt: Given $\epsilon>0$, $\exists N\in\mathbb{N}$ s.t $\int_{0}^{1} |f_n-f_m|<\epsilon$ for all $m>n>N$

$\implies \int_{0}^{1} |f_n|-\int_{0}^{1}|f_m| <\epsilon $ (Reverse triangle inequality)

But this only shows that $\int_{0}^{1} |f_n|$ is Cauchy and not $\int_{0}^{1} f_n$. Any hints would be appreciated.

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  • $\begingroup$ Presumably $f_n :[0,1] \to \mathbb{R}$? $\endgroup$ – RRL Dec 9 '18 at 1:08
  • $\begingroup$ $C[0,1]$ is the set of real-valued continuous functions, so yes! $\endgroup$ – Abe Dec 9 '18 at 1:12
  • $\begingroup$ Great -- forget about reverse triangle inequality -- $\left|\int f\right| \leqslant \int|f|$ $\endgroup$ – RRL Dec 9 '18 at 1:13
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Hint: Show $\int_0^1f_n$ forms a Cauchy sequence in $\mathbb{R}$.

$$\left| \int_0^1f_n - \int_0^1f_m\right| = \left| \int_0^1(f_n - f_m)\right| \leqslant \int_0^1 |f_n - f_m| = d(f_n,f_m)$$

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