# Convergence of the integral of a series of functions

I am having trouble proving that if $$f_k$$ converges uniformly to $$f$$ on $$[-\pi,\pi]$$ then: $$\int_{-\pi}^{\pi}f(x)dx=\lim_{k\to\infty} \int_{-\pi}^{\pi}f_k(x)dx$$

I feel like it is almost trivial, since we know that $$\forall x \in [-\pi,\pi], \forall \epsilon > 0 , \exists N \in \mathbb{N}$$ such that, if $$k \geq N \rightarrow |f_k(x)-f(x)| \leq \epsilon$$ (by uniform continuity of $$f_k$$).

Yet, I can´t work out a solution.

• Hint: There exists an $N$ s.t. $|f_k(x)-f(x)|\leq \frac{\epsilon}{2\pi}$. Subtract the two integrals and use triangle inequality. – Ninad Munshi Nov 19 '19 at 0:11
• which integrals? – PLanderos33 Nov 19 '19 at 0:22
• $\int f_k(x) dx$ and $\int f(x) dx$. Then take the absolute value of their difference, that's where the triangle inequality comes in. – Ninad Munshi Nov 19 '19 at 0:24
• Worked it out, just one last detail. Is this true? $$\int_{a}^{b} f(x) dx=|\int_{a}^{b} f(x) dx|$$ – PLanderos33 Nov 19 '19 at 0:40
• No, it is not true unless the integral was positive. – Ninad Munshi Nov 19 '19 at 0:43

HINT: the key point is that $$\left|\int_{-\pi}^{\pi }(f(x)-f_k(x))\,\mathrm d x\right|\leqslant \int_{-\pi }^{\pi }|f(x)-f_k(x)|\,\mathrm d x\tag1$$ For $$\rm(1)$$ you need to know/prove that $$|\int g|\leqslant \int |g|$$ for any integrable $$g$$ (using a convergent sequence of Riemann sums and the triangle inequality the result is immediate).
• This inequality is actually not necessary. All we need to do is show that for any $\varepsilon>0$, there exists $N$ such that for all $k\geqslant N$, we have $$\left| \int_{-\pi}^\pi f(x)\ \mathsf dx - \int_{-\pi}^\pi f_k(x)\ \mathsf dx\right| \leqslant \varepsilon.$$ – Math1000 Nov 19 '19 at 2:22