# Real Analysis: Show that function is Integrable

This question somewhat confuses me with the function being defined as $1/x-1/x$,for $(0,1]$, and 0 where x=0. I'm also somewhat thrown off about . meaning integer part. Anyway, here is my attempt, which I am not sure if is correct:

Attempt: The function is continuous at every point on [0,1]. except at x=0.
Lebesgue's Integrability Criterion states that a bounded function $f:[a,b]$ to $R$ is Riemann Integrable if and only if it is continuous almost everywhere on [a,b].
The function given is bounded on the interval [0,1]. and the point of discontinuity, x=0, is a null set, where: .

Thus, the function is continuous almost everywhere on [0,1]. and hence is Riemann integrable.

• A good attempt, but $f$ has a lot more discontinuities than just at $0$. For example, $\lim_{{x \rightarrow 1/2}^+}f(x) = 1$, but $\lim_{{x \rightarrow 1/2}^-}f(x) = 0$. Once you've actually found the set of discontinuities of $f$, the theorem you quoted will avail you. – Duncan Ramage Nov 14 '17 at 23:55
• Hello, Thanks. Why is the limit as x goes to 1/2 from the right 1? – kemb Nov 14 '17 at 23:56
• I thought it would also be 0? Thanks. – kemb Nov 14 '17 at 23:58
• @kemb Take for example $\;x=0.501\;$ , then $$\frac1x-\big\lfloor\frac1x\big\rfloor\approx1.99-1=0.99$$ and thus in the limit it equals $\;1\;$ . Try now from the left hand, say with $\;x=0.499\;$ ... – DonAntonio Nov 15 '17 at 0:03
• Oh wait, I get it now. Thank you. The brackets mean integer part. – kemb Nov 15 '17 at 0:10

Updated Answer: The function is discontinuous at 0 and points of the form $1/n$, where n is a natural number, starting from 2,3....... (I believe the function is continuous at 1, that's why we start from 2). This set is clearly countable and thus forms a null set (the set of discontinuities forms a null set).