Pointwise convergence of a sequence of piecewise functions {fn}

Question

or any x∈[0,∞), there is always Nx∈ℕ s.t. Nx>x, thus we have fn(x)=e−x,∀n≥Nx, which means

limn→∞fn(x)=e−x,∀x∈[0,∞) To show that the convergence is uniform, it suffices to show that

limn→∞supx≥0|fn(x)−f(x)|=0 For any n∈ℕ, there are 3 cases

If x≤n, then |fn(x)−f(x)|=|e−x−e−x|=0,x≤n If nn+en, then |fn(x)−f(x)|=e−x≤e−n−en≤e−n Thus we can see that

supx≥0|fn(x)−f(x)|≤max{2e−n+ne−2n,e−n}→0 as n→∞

Answer 1

You are right, if $f$ is integrable on $[a,b]$, then $F(x) = \int_a^x f(t) dt$ is continuous on $[a,b]$ and is bounded. Hence it is also integrable on $[a,b]$. Finally, as you mentioned, integrability on $[a,b]$ implies integrability on any sub-interval. Hence your claim is correct.