Question

Given \(g_n:[0,1]\to\mathbb{R}\) and \(g:[0,1]\to\mathbb{R}\), assume that there is a number \(M\) such that \(|g_n(x)\leqslant M|\) for all \(n\) and \(g(x)\leqslant M\) for all \(x\in[0,1]\). Moreover, assume that \(g_n\) and \(g\) are all integrable on \([0,1]\) and \(g_n(x)\to g(x)\) pointwise on \([0,1]\).

With the above assumptions, show that if \(g_n\to g\) uniformly on \([\delta,1]\) for any \(0<\delta<1\), then\[\lim_{n\to\infty}\int_{0}^{1}g_n=\int_{0}^{1}g.\]

Answer 1

You do not really need the last assumption to show the last limit.
The limit follows from the bounded convergence theorem. See
http://mathworld.wolfram.com/LebesguesDominatedConvergenceTheorem.html

Answer 2

May be they are giving you this assumption to make the proof easier.

Answer 3

Here is a sketch of the proof with that assumption.
\[
\int_0^{\delta} \left |g_n(t) - g(t) \right| dt\le 2M \int_0^{\delta}dt= 2 M \delta\\


\]
Let
\[
\epsilon >0
\]
and choose \[ \delta=\min(\frac { \epsilon} { 4 M}, \frac 12) \]
and choose N> so for n> M
we have
\[\left| g_n(t)- g(t) \right| \le \frac \epsilon {4 (1-\delta)}
\]
for any t in
\[ [ \delta, 1]\le\\

\]
\[
\int_0^{1} \left |g_n(t) - g(t) \right| dt=\int_0^{\delta} \left |g_n(t) - g(t) \right|dt+
\int_{\delta}^1 \left |g_n(t) - g(t) \right|dt\le2 M \delta + \frac \epsilon {4 (1-\delta)}(1-\delta)\\
\frac \epsilon 2 + \frac \epsilon 4 < \epsilon
\]

Answer 4

abs Int <int abs so it is going to zero.