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.
$\endgroup$ 24 Answers
$\begingroup$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).
$\endgroup$ $\begingroup$If you refer $\int_{(0,\,1)} y \,dx$ to the Riemann integral of $y: (0,\,1) \to \mathbb{R}$ over $(0,\,1)$, there is no such a thing. Since Riemann integral is only defined on a closed interval. But if we refer the case to general integral, it works for Lebesgue integral, and it has the same value as $\int_{[0,\,1]} y \,dx$, as we will see below.
Let me invoke measure theory to answer the question. First, suppose $y: [0,\, 1] \to \mathbb{R}$ is Riemann integrable. In measure theory, every Riemann integrable function is a Lebesgue integrable function. And a function $f: X \to \mathbb{R}$ is Lebesgue integrable if and only if$$ \int_X|f| \,d\mu < \infty \,. $$And of course, $f$ must be measurable (i.e., for every measurable set $A \subseteq \mathbb{R}$, $f^{-1}(A)$ is measurable on $X$). Hence $y$ is Lebesgue integrable by definition.
And note that every singleton in $\mathbb{R}$ has Lebesgue measure zero, i. e., $\mu(\{x\}) = 0$ for every $x \in \mathbb{R}$. Then we also have $[0,\, 1] = (0,\, 1) \cup \{0, 1\}$ and $(0,\, 1) \cap \{0, 1\} = \varnothing$, which means that $\{(0,\, 1),\, \{0, 1\}\}$ is a partition of $[0,\, 1]$.
Then by simple function approximation theorem, for every $k \in \mathbb{N}$, there exist measurable simple functions $s_k^+, s_k^-: [0,\, 1] \to [0,\, 1)$ such that$$ \forall x \in [0,\, 1]\,\big[ 0 \leq s_1^+(x) \leq s_2^+(x) \dotsb \leq y^+(x) \big] \;\land\; \forall x \in [0,\, 1] \left[ \lim_{k \to \infty} s_k^+(x) = y^+(x) \right] $$and$$ \forall x \in [0,\, 1]\,\big[ 0 \leq s_1^-(x) \leq s_2^-(x) \dotsb \leq y^-(x) \big] \;\land\; \forall x \in [0,\, 1] \left[ \lim_{k \to \infty} s_k^-(x) = y^-(x) \right] $$where $y^+ := \max \{y, 0\}$ and $y^- := \max \{-y, 0\}$. Note that $y^+$ and $y^-$ are measurable since $y$ is Lebesgue integrable. Then for every $k \in \mathbb{N}$and $* \in \{+, -\}$, let $m_k^* \in \mathbb{N}$, and we define$$ \mathrm{image}(s_k^*) := \{\alpha_{(k, 1)}^*, \dotsc, \alpha_{(k, m_k)}^*\} $$and$$ \forall j \in \{1, \dotsc, m_k^*\} \,,\; A_{(k,\, j)}^* := (s_k^*)^{-1}(\{\alpha_{(k,\, j)}^*\}) \,, $$and then we can express $s_k^*$ by$$ s_k^* = \sum_{j = 1}^{m_k^*} \alpha_{(k, j)}^* \chi_{A_{(k, j)}^*} \,, $$where $\chi$ is an indicator function. Note that $\left\{A_{(k, j)}^*\right\}_{j = 1}^{m_k^*}$ forms a partition of $[0,\,1]$, for every $k \in \mathbb{N}$ and $* \in \{+, -\}$. And since $A_{(k, j)}^* \cap \{0\} \subseteq \{0\}$ and $A_{(k, j)}^* \cap \{1\} \subseteq \{1\}$, hence we have$$ 0 \leq \mu(A_{(k, j)}^* \cap \{0\}) \leq \mu(\{0\}) = 0 $$and$$ 0 \leq \mu(A_{(k, j)}^* \cap \{1\}) \leq \mu(\{1\}) = 0 \,, $$which show that $\mu(A_{(k, j)}^* \cap \{0\}) = 0$ and $\mu(A_{(k, j)}^* \cap \{1\}) = 0$, for every $j \in \{1, \dotsc, m_k^*\}$.
Then by the property of Lebesgue integral we obtain\begin{align} \int_{[0,\, 1]} y\,d\mu &= \int_{\{0\} \cup (0,\, 1) \cup \{1\}} y \,d\mu \\ &= \int_{\{0\}} y\,d\mu + \int_{(0,\, 1)} y \,d\mu + \int_{\{1\}} y \,d\mu \\ &= \left[ \int_{\{0\}} y^+ \,d\mu - \int_{\{0\}} y^- \,d\mu \right] + \left[ \int_{(0,\,1)} y^+ \,d\mu - \int_{(0,\,1)} y^- \,d\mu \right] + \left[ \int_{\{1\}} y^+ \,d\mu - \int_{\{1\}} y^- \,d\mu \right] \\ &= \left[ \sup_{k \in \mathbb{N}} \int_{\{0\}} s_k^+ \,d\mu -\sup_{k \in \mathbb{N}} \int_{\{0\}} s_k^- \,d\mu \right] + \left[\sup_{k \in \mathbb{N}} \int_{(0,\,1)} s_k^+ \,d\mu - \sup_{k \in \mathbb{N}} \int_{(0,\,1)} s_k^- \,d\mu \right] \\&\quad\quad + \left[\sup_{k \in \mathbb{N}} \int_{\{1\}} s_k^+ \,d\mu - \sup_{k \in \mathbb{N}} \int_{\{1\}} s_k^- \,d\mu \right] \\ &= \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \mu\left(A_{(k, j)}^+ \cap \{0\}\right)\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \mu\left(A_{(k, j)}^- \cap \{0\}\right)\right) \right] \\ &\quad\quad+ \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \mu\left(A_{(k, j)}^+ \cap (0,\,1)\right)\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \mu\left(A_{(k, j)}^- \cap (0,\,1)\right)\right) \right] \\ &\quad\quad+ \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \mu\left(A_{(k, j)}^+ \cap \{1\}\right)\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \mu\left(A_{(k, j)}^- \cap \{1\}\right)\right) \right] \\ &= \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \cdot 0\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \cdot 0\right) \right] \\ &\quad\quad+ \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \mu\left(A_{(k, j)}^+ \cap (0,\,1)\right)\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \mu\left(A_{(k, j)}^- \cap (0,\,1)\right)\right) \right] \\ &\quad\quad+ \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \cdot 0\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \cdot 0\right) \right] \\ &= (0 - 0) + \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \mu\left(A_{(k, j)}^+ \cap (0,\,1)\right)\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \mu\left(A_{(k, j)}^- \cap (0,\,1)\right)\right) \right] \\ &\quad\quad +(0 - 0) \\ &= \left[ \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^+} \alpha_{(k, j)}^+ \mu\left(A_{(k, j)}^+ \cap (0,\,1)\right)\right) - \left(\sup_{k \in \mathbb{N}} \sum_{j = 1}^{m_k^-} \alpha_{(k, j)}^- \mu\left(A_{(k, j)}^- \cap (0,\,1)\right)\right) \right]\\ &= \sup_{k \in \mathbb{N}} \int_{(0,\,1)} s_k^+ \,d\mu - \sup_{k \in \mathbb{N}} \int_{(0,\,1)} s_k^- \,d\mu \\ &= \int_{(0,\,1)} y^+ \,d\mu - \int_{(0,\,1)} y^- \,d\mu \\ &= \int_{(0,\,1)} y \,d\mu \,. \end{align}
Recalling that $y$ is also Riemann integrable, then the Lebesgue integral is equal to the Riemann integral. And hence we obtain$$ \int_{(0,\,1)} y \,d\mu = \int_{[0,\,1]} y \,d\mu = \int_0^1 y \,dx \,. $$
I hope this answer could help. $\Box$
$\endgroup$ $\begingroup$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 $\delta(\infty) \cdot e^{\infty}$
$\endgroup$ $\begingroup$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 $f(1)$ )
$\endgroup$
