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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.

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

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  • $\begingroup$ 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 $$ \left| \int_{-\pi}^\pi f(x)\ \mathsf dx - \int_{-\pi}^\pi f_k(x)\ \mathsf dx\right| \leqslant \varepsilon. $$ $\endgroup$ – Math1000 Nov 19 '19 at 2:22

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