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Let $f, f_n : [a,b] \to \mathbb{R} $ be Riemann $\forall n \in \mathbb{N}$ if $\lim_{n \to \infty} \sup _x |f_n(x) - f(x)| = 0$, then $\lim _{n \to \infty} \int_a^b f_n = \int_a^b f$

Note: I don't have uniform convergence at my disposal.

Let $ \varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $\forall n \ge N$ $\sup_{x \in [a,b]} \lvert f_n - f \rvert < \varepsilon$ Since $\forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = \max\{| \sup f_n|, : n \in \mathbb{N}\} \cup \{|\sup f|\}$

$\lvert \int_a^b f_n - f \rvert \le \int_a^b \lvert f_n-f| \le 2M(b-a)$.

I'm not sure where to use the supremum at I guess is my issue.

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    The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $\int_a^b g(x)\,dx\leq (b-a)\sup\limits_{a\leq x\leq b}g(x)$. – 2017-01-31
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    @JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent. – 2017-01-31
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    I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly. – 2017-01-31
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    @JonasMeyer yeah thats exactly what I mean. Sorry for the confusion. – 2017-01-31

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Consider $\epsilon>0$

Then from $\lim_{n\to\infty}\sup_x|f_n(x)-f(x)|=0$ you can find a $N\in\mathbb N$ such that $\sup_x|f_n(x)-f(x)|<\frac{\epsilon}{b-a}$, for all $n\ge N$.

Which means $|f_n(x)-f(x)|<\frac{\epsilon}{b-a}\ \ \forall x\in[a,b]\ \ \forall n\ge\mathbb N$

Then:

$$\left|\int_a^b f_n-\int_a^bf\right|\le \int_a^b |f_n-f|\le\frac{\epsilon}{b-a}(b-a)=\epsilon$$

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    if $sup_x \lvert f_n(x) - f(x) \rvert < \frac{\varepsilon}{b-a} $ then $\lvert f_n(x) - f(x) \rvert< \frac{\varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser) – 2017-01-31
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    All elements of a set are smaller than its supremum, which is $x\le \sup A$ for all $x\in A$ – 2017-01-31
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    In other words, $\lvert f_n(x) - f(x) \rvert \le sup_x \lvert f_n(x) - f(x) \rvert$ ? – 2017-01-31
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    Yes, that's right. – 2017-01-31