# Show that $\lim_{n\to\infty} \int_{0}^{1} f_n$ exists

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.

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

Hint: Show $$\int_0^1f_n$$ forms a Cauchy sequence in $$\mathbb{R}$$.
$$\left| \int_0^1f_n - \int_0^1f_m\right| = \left| \int_0^1(f_n - f_m)\right| \leqslant \int_0^1 |f_n - f_m| = d(f_n,f_m)$$