Let be integrable on [0,1] for all and uniformly. Show that is integrable and .
Since converges uniformly to , for any , there is such that for all and for all , . And since , .
So .
i have a question here. i know that for each , but how can i show that .
help would be appreciated so much.
You answered yourself. To cover it the way I said remember how one constructed the set by taking intervlas of lenght (if I rememeber correctly) and then intersecting all of them so it suffices to cover a "step" for sufficiently large .
The fact that if such a cover exists then is integrable is more or less easy to see.
Well, I guess you also need to have f bounded on the hypothesis, but given it's Riemann it's always implicitly assumed.