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Let $f$ be defined as:
$$ f(x) = \begin{cases} x & \text{ if } x\in\mathbb{Q}; \\\\ 0 & \text{ if }x\notin\mathbb{Q}. \end{cases} $$ Is $f$ Riemann integrable on $[0,1]$? Prove it.

We know that the upper sum $U = x$ and lower sum $L=0$.

Since $\lim U\neq \lim L$, $f(x)$ is not Riemann integrable.

Is this sufficient for the proof?

user606854
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1 Answers1

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Start from the basic. Divide the interval [0,1] into $n$ sub-intervals, each of length $\frac{1}{n}$, i.e. $[0,\frac{1}{n}], [\frac{1}{n}, \frac{2}{n}], \cdots, [\frac{n-1}{n}, 1]$, Note that, at each sub-interval of the form $[\frac{i}{n}, \frac{i+1}{n}]$, maximum value of the function = $\frac{i+1}{n}$ and minimum value is 0, as each sub-interval has at-least one rational and one irrational number. Then show that Lower integral converges to 0, but upper integral does not which will show that the function is not Riemann Integrable.

Debarghya
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