Show that if a sequence of integrable functions converges uniformly on [a,b] to

Question

Show that if a sequence of integrable functions converges uniformly on [a,b] to a function f then f is also integrable

Explanation / Answer

Suppose f_n ---> f uniformly, and f_n is integrable for each n. I want to now show that: U(f_n) ---> U(f) L(f_n) ---> L(f) Since f_n is integrable, U(f_n) = L(f_n) for all n, so by uniqueness of limits, this would imply that U(f) = L(f), i.e. f is integrable. Also, just a quick note on terminology: if I write "partition" in this answer, I am referring to a cover of the interval [a, b] by closed subintervals with disjoint interiors. This is not technically a partition in the usual sense, since the parts would normally be disjoint, not just their interiors. Fix e > 0. We can choose partitions P'_n and P such that the upper sum of f_n over P'_n is less than e/3 from U(f_n), and the upper sum of f over P is less than e/3 from U(f). Now, by letting, for each n: P_n = {X intersct Y : X is in P'_n and Y is in P} we form another partition which is finer than both P'_n and P, and hence the upper sums of f_n and f over P_n will be within e/3 of U(f_n) and U(f) respectively. Now, since f_n ---> f uniformly, there exists some N such that, for all x in [a, b]: n > N ===> |f_n(x) - f(x)| N, the rectangle heights of the upper sum for f_n will be within e / [3(b - a)] of the rectangle heights of f. In total, this difference can be at most: (b - a) * e / [3(b - a)] = e/3 So, for n > N, U(f_n) is within e/3 of the upper sum over P_n of f_n, which is within e/3 of the upper sum of f over P_n, which is within e/3 of U(f). Therefore: n > N ===> |U(f_n) - U(f)| U(f). Proving the same for L(f_n) ---> L(f) is almost exactly the same. In fact, we could prove it by replacing f with -f. Therefore, L(f) = U(f), so f is integrable

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