- #1
bbkrsen585
- 10
- 0
Suppose gn are nonnegative and integrable on [0, 1], and that gn \rightarrow g almost everywhere.
Further suppose that for all \epsilon > 0, \exists \delta > 0 such that for all A \subset [0, 1], we have
meas(A) < \delta implies that supn \intA |gn| < \epsilon.
Prove that g is integrable, and that \int[0,1] g = lim \int[0,1] gn.
Attempt at the solution:
There are some observations I've compiled. First, for an set A with measure less than \delta, we know that the the limiting function g exists on that set. So, it's integral is also bounded by the sup. But from here, I don't know how to argue that the limiting function has a uniform bound on all of [0, 1].
Further suppose that for all \epsilon > 0, \exists \delta > 0 such that for all A \subset [0, 1], we have
meas(A) < \delta implies that supn \intA |gn| < \epsilon.
Prove that g is integrable, and that \int[0,1] g = lim \int[0,1] gn.
Attempt at the solution:
There are some observations I've compiled. First, for an set A with measure less than \delta, we know that the the limiting function g exists on that set. So, it's integral is also bounded by the sup. But from here, I don't know how to argue that the limiting function has a uniform bound on all of [0, 1].
