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What is one difference in the values of

$$\int\limits_{\left[0,1\right]}y\, dx$$$$\int\limits_{\left(0,1\right)}y\, dx$$

and how would you calculate the values? For the sake of simplicity, let $y=x$. Conceptualizing integration as the area bounded by the function, the $x$-axis and the limits of integration, the latter should be smaller.

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    $\begingroup$ There is no difference. Usual Riemann-Lebesgue integration does not see these points as long as they are not atoms of your measure. $\endgroup$ – b00n heT Oct 22 '16 at 21:29
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    $\begingroup$ Texts generally define Riemann integration on closed intervals. This is really a definitional convention, since defining it on open intervals amounts to the same thing. $\endgroup$ – MathematicsStudent1122 Oct 22 '16 at 21:55
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It should be intuitive that $\displaystyle \int_{(0, 1)} f(x) \ dx = \int_{[0, 1]} g(x) \ dx$ where $g(x) = \begin{cases} f(x) & \ \text{ if }\ x \in (0, 1) \\ 0 & \ \text{ if } \ x \in \{0, 1\}\end{cases}$. We claim that $\displaystyle \int_{[0, 1]} f(x) \ dx = \int_{[0, 1]} g(x) \ dx$, or more generally, changing the value of $f$ at finitely many points has no effect on the value of the definite integral.


Sketch of proof:

Provided a function $f$ is integrable on an interval $[a, b]$, the definite integral is rigorously defined as follows: there is a unique $I$ such that, for any given partition $\mathcal{P}$ of an interval $[a, b]$, we have:

$$L(f, \mathcal{P}) \leq I = \int_a^b f(x) \ dx \leq U(f, \mathcal{P})$$

Where $\displaystyle L(f, \mathcal{P}) = \sum_{i} (x_{i+1} - x_i)\inf \Big( \{f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$ where $x_i$'s $\in \mathcal{P}$

and likewise $\displaystyle U(f, \mathcal{P}) = \sum_i (x_{i+1} - x_i)\sup \Big( \{ f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$

Now suppose we change the value of $f$ at a point $y \in [a, b]$. For any given partition, we can "refine" this partition to encapsulate $y$ inside an arbitrarily small interval, in effect making its associated term in the $L(f, \mathcal{P}')$ and $U(f, \mathcal{P}')$ summations arbitrarily insignificant (limiting to zero in successive such refinements of the partition).

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This is not a definitive answer, considering an answer have already accepted almost 2 years ago, but I believe it should be noted that in the case where there exists a Dirac delta function or any similarly defined special functions inside the integrand, the integral over some 'points' become non-zero by definition.

Using this fact, we can prove that the claim that these two integrals are equal leads to a contradiction if one or both of the limits of the interval are included a specially defined function. In this case, inclusion of a particular point will change the answer of the integration by a finite amount. Choosing this point from the limits of the interval, integration over an open interval will yield a different result than integration over an open interval by that finite amount.

So, integration on open or closed intervals can yield different results if there is a special function in the integrand such as Dirac delta.


Disclaimer: Notion of integrating δ(x) over an interval [0,a] might be undefined on its own. If anybody have any information in this regard, please edit the answer accordingly.


See also: Dirac delta integral with $\delta(\infty) \cdot e^{\infty}$

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There is not difference. $$\int_{[0,1]}f(x).dx = \int_{(0,1)}f(x).dx = \lim_{n\to \infty} \left(I_n^+(f,0,{1\over 2}) + I_n^-(f,{1\over 2},1)\right) $$

With $I_n^+$ and $I_n^-$ respectively the upper and lower Riemann sums ( I chose such a decomposition because those Riemann sums never involve $f(0)$ and $f(1)$ )

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