# Integration over an open interval.

Is there any difference between the following integrations: $$\int_a^b f(x) dx$$ where $$(a,b]$$

And $$\int_a^b f(x) dx$$ where $$[a,b]$$

• Do you know a bit of measure theory? Dec 7, 2023 at 16:28
• @SineoftheTime, yes Dec 7, 2023 at 20:59
• basically, if you remove a point you're removing a set of measure zero so the value of the integral does not change Dec 7, 2023 at 21:05
• would you recommend a reference ? Dec 7, 2023 at 21:30
• What kind of integrals? Where did this question came from? Needs context. Dec 7, 2023 at 21:47

In the context implicit in your question the answer is "no". In the most elementary terms, the definite integral is the limit of the sums of areas of sets of rectangles that approximate the area under the graph of $$f$$. That area does not change if you change the value of $$f$$ at any finite number of points, or omit the area "above" any finite number of points.

This should also answer your other question at For an integration of a closed interval , how can we exclude a certain point from this integration?

• would you please recommend a reference Dec 7, 2023 at 21:02
• The answer from @pie has a proof and a reference. Dec 8, 2023 at 1:10

This theorem comes from Introduction to Real Analysis Book by Robert G. Bartle 7.1.3

For $$a\leq b$$ let the $$x_0=a ,x_n=b$$ choose $$n-1$$ points in $$[a,b]$$ and call them $$x_2 ,x_3, \dots x_{n-1}$$ such that $$x_k > x_{k-1}$$ then make the interval $$I_k:= [x_{k-1}, x_k]$$ for $$0 \leq k \leq n$$ define the norm of that partition by $$|P|= \sup\limits_{k=1}^n \{ x_{k}- x_{k-1} \}$$, from every interval $$I_k$$ choose any point and label it $$t_k$$ then we can define the integral $$\int_a ^b f(t) dt$$ if $$\forall \epsilon >0 \ \exists \delta>0$$ such that if $$|P|<\delta$$ then $$\left| \sum\limits_{k=1}^n f(t_k) (x_{k}- x_{k-1}) - L \right| < \epsilon$$ for some $$L \in \mathbb{R}$$ that number $$L$$ is $$\int_a ^b f(t) dt$$ if $$a> b$$ we define $$\int_a ^b f(t) dt=-\int_b ^a f(t) dt$$ we call $$f$$ a Riemann intergable function

Theorem If $$g$$ is Riemann integrable on $$[a, b]$$ and if $$f (x) =g (x)$$ except for a finite number of points in $$[a, b]$$, then $$f$$ is Riemann integrable and $$\int _a^b f(x)dx= \int_a^b g(x)dx$$

proof Let $$c$$ be a point in the interval and let $$L:= \int_a^b g(x)dx$$. Assume that $$f(x)=g(x)$$ for all $$x\ne c \in[a,b]$$. For any tagged partition $$P$$ , the terms in the two sums $$\sum\limits_{k=1}^n f(t_k) (x_{k}- x_{k-1}) , \sum\limits_{k=1}^n g(t_k) (x_{k}- x_{k-1})$$ are identical with the exception of at most two terms (in the case that $$c$$ is an endpoint). Therefore, we have $$\left|\sum\limits_{k=1}^n f(t_k) (x_{k}- x_{k-1}) - \sum\limits_{k=1}^n g(t_k) (x_{k}- x_{k-1})\right|\leq 2|P|(|f(c) |+|g(c)|)$$

given $$\varepsilon>0$$ since $$g$$ is a is Riemann integrable on $$[a, b]$$ choose $$\delta_1$$ st if $$|P| < \delta_1$$ implies $$\left| \sum\limits_{k=1}^n g(t_k) (x_{k}- x_{k-1}) -L\right| <\frac \varepsilon 2$$ and choose $$\delta_2 =\frac{\varepsilon}{8(|f(c) |+|g(c)|)}$$ and choose $$\delta = \min \{\delta_1, \delta_2 \}$$

$$\left| \sum\limits_{k=1}^n f(t_k) (x_{k}- x_{k-1})-L\right|\leq \left| \sum\limits_{k=1}^n g(t_k) (x_{k}- x_{k-1})-L\right| + \left| \sum\limits_{k=1}^n f(t_k) (x_{k}- x_{k-1})-\sum\limits_{k=1}^n g(t_k) (x_{k}- x_{k-1})\right| < \varepsilon$$

• is this theorem valid for infinit integral?? Dec 17, 2023 at 21:24
• @ElmarconiAbdo do you mean improper integrals?
– pie
Dec 18, 2023 at 5:22
• Yes. Does the same theory applicable for improper integrals? Dec 18, 2023 at 14:57
• Thanks for your qualified response, I am from Egypt. May I ask for getting in contact with you? Dec 20, 2023 at 4:56
• صباح الورد من أم الدنيا Dec 20, 2023 at 6:57