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Let $\{x_j\}_{j \in \mathbb{N}} $ be the set of rational numbers in $\mathbb{Q}\cap [0,1]$, and let $f_n(x) = 1_{\{x_j , \ j \leq n\}}$ i.e $f_n$ is $1$ if $x$ lies in the set $\{x_j\}_{0 < j \leq n}$ otherwise.

Why is $\int_0^1 f_n(x) dx = 0$?

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    Isn't $f_n$ zero except at a finite number of points? – 2017-02-22
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    It clearly is non-negative, so you need to show that it is smaller than $\epsilon$ for any $\epsilon>0$. – 2017-02-22
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    Note that $f_n(x)=\sum_{i=1}^{n} 1_{\{x_i\}}$. So you only really need to prove that $\int_{0}^{1} 1_{\{x_0\}}(x)\,dx=0$ for any $x_0$. – 2017-02-22
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    Yes, that's correct, can you elaborate? @MPW – 2017-02-22

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Given any partition $P$, each subinterval $I$ contains an irrational number where $f_n(x) = 0$. Hence, $\inf_{x \in I} f_n(x) \leqslant 0 \leqslant \sup_{x \in I} f_n(x)$.

Forming the upper and lower sums we have

$$L(P,f_n) \leqslant 0 \leqslant U(P,f_n).$$

Hence, upper and lower integrals satisfy

$$\underline{\int}_0^1 f_n = \sup_{P} L(P, f_n) \leqslant 0 \leqslant \inf_{P} \,U(P,f_n) = \overline{\int}_0^1f_n.$$

Since $f_n$ is continuous almost everywhere it is Riemann integrable and the upper and lower integrals must be equal. Thus,

$$\underline{\int}_0^1 f_n = \overline{\int}_0^1f_n = \int_0^1 f_n = 0.$$