Uniform continuity problem.

Question

Let $\{f_n(x)\}$ a sequence of real-valued continuous functions defined on $[0,1]$ and suppose that $f_n$ converges uniformly to a function $f$ on $[0,1]$. Show that $$\lim_{n\rightarrow\infty}\int_{0}^{1-1/n}{f_n(x)dx}=\int_{0}^{1}{f(x)dx}$$

Answer 1

Since $f$ is continuous, we can fix $M$ such that $|f(x)| \leq M$ on $[0, 1]$. We have the expression \begin{align*} \left| \int_{0}^{1} f(x) \mathrm{d} x - \int_{0}^{1 - 1/n} f_{n}(x) \mathrm{d}x \right| & = \left| \int_{0}^{1} (f(x) - f_{n}(x)) \mathrm{d} x + \int_{1 - 1/n}^{1} f(x) \mathrm{d} x \right| \\ & \leq \int_{0}^{1} |f(x) - f_{n}(x) | \mathrm{d}x + \int_{1 - 1/n}^{1} |f_{n}(x)| \mathrm{d} x \\ \leq (1) \| f - f_{n} \| + (1/n) \| f_n \| , \end{align*} where $\| g \| = \max \{ |g(x)| : 0 \leq x \leq 1 \}$.

Now fix $\epsilon > 0$, and let $n$ be sufficiently large that $\| f - f_n \| < \min \{ \epsilon / 2, 1 \}$, and $(M + 1)/n < \epsilon / 2$. Then $\| f_{n} \| |leq \| f_n - f \| + \| f \| \leq 1 + M$. So \begin{align*} \| f - f_{n} \| + (1/n) \| f_n \| & < \epsilon / 2 + (M + 1)/n \\ & \epsilon / 2 + \epsilon / 2 \\ & = \epsilon . \end{align*}