Question

construct a sequence of a function (fn) such that each (fn) is not Riemann integrable but it's uniform limit is Riemann integrable​

Answer 1

Answer:

hope it's helpful to you!

Step-by-step explanation:

the second one doesn't hold assuming you're talking about RR with usual measure. take ff to be a function s.t. it's 00 for negative xx and 1n1n on the intervals [n−1,n][n−1,n] for positive integer nn. then ff is not integrable because ∑1n∑1n is divergent, while taking fnfn's to be ff 'cut out' after nn, so fn(x)=f(x)fn(x)=f(x) for x≤nx≤n and fn(x)=0fn(x)=0 for x>nx>n yields uniform convergence adn all fnfn's are integrable

also you can make this counterexample work for the continuous case as well by slightly regularizing your functions, so perhaps you're missing some assumptions?

edit: for the edited version of your question you just take nn large enough so that |fn−f|<1|fn−f|<1 for all xx and then you use |f|≤|fn|+|fn−f||f|≤|fn|+|fn−f| - it's triangle inequality