Showing that $\lim_{n \to \infty} \int_0^1 g_n = \int_0^1 g.$

Let $g_n$ and $g$ be uniformly bounded on $[0,1]$, meaning that there exists a single $M > 0$ satisfying $|g(x)| \leq M$ and $|g_n(x)| \leq M$ for all $n \in \mathbb{N}$ and $x \in [0,1]$. Assume $g_n \rightarrow g$ pointwise on $[0,1]$ and uniformly on any set of the form $[0, \alpha]$ where $0 < \alpha < 1$. If all the functions are (Riemann) integrable, show that $\lim_{n \to \infty} \int_0^1 g_n = \int_0^1 g.$

I tried proving this, but I'm not sure if my reasoning is entirely correct. Let $\epsilon > 0$. Fix $\alpha \in (0,1)$ and consider the interval $[0, \alpha]$. Then $$\left| \int_0^1 g_n - \int_0^1 g \right| \\ = \left| \int_0^{\alpha} g_n + \int_{\alpha}^1 g_n - \left( \int_0^{\alpha} g + \int_{\alpha}^1 g \right) \right| \\ \leq \int_{0}^{\alpha} | g_n(x) - g(x) | dx + \int_{\alpha}^1 |g_n(x) - g(x)| dx \\ \leq \int_{0}^{\alpha} | g_n(x) - g(x) | dx + \int_{\alpha}^1 |g_n(x)| dx + \int_{\alpha}^1 |g(x)| dx.$$ Since $g_n \rightarrow g$ uniform on $[0, \alpha]$, there exists a $n_0 \in \mathbb{N}$ such that $$|g_n(x) - g(x)| < \frac{\epsilon}{2 \alpha},$$ for all $x \in [a,b]$ and all $n \geq n_0$. Since $g_n$ and $g$ are uniformly bounded, we have $|g_n(x)| < M$ and $|g(x)| < M$. So all together this would give me $$\left| \int_0^1 g_n - \int_0^1 g \right| < \frac{\epsilon}{2\alpha} + 2M(1-\alpha).$$ How can I get this smaller than $\epsilon$? Also, I'm not sure if my reasoning is correct since I haven't used the fact that $g_n \rightarrow g$ pointwise on $[0,1]$.

Help is appreciated.

• Are your functions continous ?
– yago
Aug 5, 2016 at 12:52
• That is not given. All is given, is that they are Riemann integrable. Aug 5, 2016 at 12:54
• If $|g_n(x) - g(x)| \leq \frac{\epsilon}{2\alpha}$ then $\int_0^\alpha|g_n(x) - g(x)|dx \leq \frac{\epsilon}{2}$, not $\frac{\epsilon}{2\alpha}$. Then your bound is $\frac{\epsilon}{2} + 2M(1-\alpha)$, so by choosing $\alpha$ small enough, you're done
– yago
Aug 5, 2016 at 12:59
• Why do you still have $\alpha$ in the denominator after "So all together this would give me"? And since you can choose the $\alpha$ you want to use, to control the second term have you, for instance, tried to set $\alpha\stackrel{\rm def}{=} 1-\frac{\varepsilon}{4M}$? Aug 5, 2016 at 13:00
• (Also, the pointwise convergence does not seem necessary, to answer your doubt. Note that the uniform convergence you get imply already $f_n\to f$ pointwise on $[0,1)$; and the convergence pointwise at $1$ is not useful nor necessary for the result you want.) Aug 5, 2016 at 13:02

Hint: You're so close. Take $\alpha$ close to $1$. Then we can bound the difference of the integrals on $[0, \alpha]$ since it is uniform. Then we can bound the difference of the integrals by $2M(1 - \alpha)$, as you have done: since $\alpha$ is close to $1$, this quantity will also be small.
• Can you be more specific please. Like write down explicitly the last steps of the argumentations? Since I still don't know how to pick $\alpha$ to get the correct upper bound. Aug 5, 2016 at 13:05
• Let $1 - \alpha < \frac{\varepsilon}{2M}$. Then (Using your last line) we have $\frac{\varepsilon}{2\alpha} + M(1 - \alpha) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$.
• @MichaelTong You are missing a factor $2$ in the last line. (Also, there shouldn't be any $\alpha$ in the denominator.) Aug 5, 2016 at 13:07
• @ClementC. You mean in the answer or the comment? In the comment I was just leaving off where Kamil started, though I agree there shouldn't be an $\alpha$ in the denominator of $\frac{\varepsilon}{2 \alpha}$