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Let $f:[0,2] \to R$ defined as $f(x)= \begin{cases} 1 & x \neq 1 \\ 0 & x = 1 \end{cases}$

Prove that $f$ is integrable in $[0,2]$ and obtain its integral.

Well since the function it's not continuous nor monotonic, then I think this must be proved by definition, but I don't know how to find the partitions, upper and lower sum, etc.

aras
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MarioM21
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    Every function with a finite number of discontinuities is Riemann integrable. Idea: Split your interval such that the discontinuities are on the boundary of the partition intervals. – Benjamin Nov 28 '20 at 19:13
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    One can even prove this holds, if the set of discontinuities is countable (cf. https://math.stackexchange.com/questions/263189/proof-that-a-function-with-a-countable-set-of-discontinuities-is-riemann-integra). – Benjamin Nov 28 '20 at 19:15
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    I edited your question to make your function definition for $f(x)$ more readable using begin{cases}...end{cases}. This might be useful to you in the future. – aras Nov 28 '20 at 19:17
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    Note that $f$ is piecewise continuous in [0,2] and hence integrable. To obtain the integral, simply break the integral at the point $1$ and calculate. – Nik Nov 28 '20 at 19:21
  • Thanks a lot aras and @Benjamin, so I can split the interval into the partitions $P_1=[0,1]$ and $P_2=[1,2]$ since $x=1$ is a point of discontinuity? – MarioM21 Nov 28 '20 at 19:29

1 Answers1

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hint

Let $ \epsilon>0 $ small enough.

Consider the partage $$P=(0,1-\frac{\epsilon}{3},1+\frac{\epsilon}{3},2)$$

then

$$U(f,P)-L(f,P)=$$ $$(1-1)(\frac{1-\epsilon}{3}-0)+$$ $$(1-0)(1+\frac{\epsilon}{3}-(1-\frac{\epsilon}{3}))+(1-1)(2-(1+\frac{\epsilon}{3}))=$$ $$\frac{2\epsilon}{3}<\epsilon$$

Its integral is given by

$$\int_0^1f+\int_1^2f=$$ $$\lim_{n\to+\infty}(\int_0^{1-\frac 1n}dx+\int_{1+\frac 1n}^2dx)=$$ $$\lim_{n\to+\infty}(1-\frac 1n)+(2-1-\frac 1n)=$$ $$\lim_{n\to+\infty}(2-\frac 2n)=2$$m

hamam_Abdallah
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