[Math] If $f$ is zero except at finitely many points, then $\int_{a}^{b} f = 0$

proof-verificationreal-analysisriemann-integration

Let $f: [a,b] \to \mathbb{R}$ be zero everywhere except at the points $s_1, \dots s_k$. Prove that $f$ is integrable and $\int_{a}^b f = 0$, directly from the definition.

Attempt: Let $\epsilon > 0$. Let $M:= \max_{i=1}^k |f(s_i)|$.

Put $\delta = \epsilon/(Mk)$. Let $P= (x_0, \dots x_n)$ be a partition of $[a,b]$ with $\Vert P \Vert < \delta$ and let $T = (t_1, \dots t_n)$ be points with $t_i \in [x_{i-1},x_i]$.

Then $|S(f,P,T)-0| \leq \sum_{i=1}^n |f(t_i)|(x_i – x_{i-1}) \leq M k \Vert P \Vert < \epsilon$

Is this correct?

Best Answer

I suppose that you want to consider the Riemann approach to the integral of one-dimensional real functions. I just want to give a glimpse on the approach to this via the regulated integral. Although this is probably not what you're looking for, hopefully it is of some interest for you.

I try to outline the concept of the regulated integral in the following. If you are familiar with it, you can just skip ahead to see why this approach is very fitting for your problem.

Definition: Let $a<b$. A finite partition of $[a,b]$ is a set $P=\{x_0,\dots,x_n\}\subseteq [a,b]$ s.t. $$a=x_0<x_1<\dots<x_{n-1}<x_n=b$$

First, we define a class of functions, so called step-functions:

Definition: A function $f:[a,b]\to\mathbb{R}$ is called a step function if there exists a finite partition $P=\{x_0,\dots,x_n\}$ of $[a,b]$ s.t. $$f|_{(x_{j-1},x_j)}:(x_{j-1},x_j)\to\mathbb{R}, x\mapsto f(x)$$ is constant for every $j\in\{1,\dots,n\}$. One calls such a finite partition fitting to the step function $f$.

Note, that as $f$ is constant on these open intervals induced by $P$, there exists constants $c_1,\dots, c_n\in\mathbb{R}$ s.t. $f|_{(x_{j-1},x_j)}(x)=c_j$ for all $x\in (x_{j-1},x_j)$ and $j\in\{1,\dots,n\}$.

For a step function, we now define the definite integral as the sum of the areas of the squares induced by the steps.

Definition: Let $f$ be a step function with fitting finite partition $P=\{x_0,\dots,x_n\}$. Then $$\int_Pf=\sum_{j=1}^nc_j(x_j-x_{j-1})$$ is the integral of $f$ over $P$.

This definition is invariant under change of partition:

Theorem: Let $f:[a,b]\to\mathbb{R}$ be a step function and $P,P'$ be two fitting finite partitions of $[a,b]$. Then $$\int_Pf=\int_{P'}f$$ and followingly we define $$\int_a^bf=\int_Pf$$ for some(any) fitting finite partition of $[a,b]$ as the definite integral of the step function $f$.

The regulated integral is now defined over a class of functions $f$ which can be represented as the uniform limit of a function sequence $f_n$ of step functions. This is collected in the following theorem:

Theorem: Let $f:[a,b]\to\mathbb{R}$ be a function such that $$\lim_{n\to\infty}f_n=f\text{ uniformly}$$ for a sequence of step functions $(f_n)_n$. Then $$a=\lim_{n\to\infty}\int_{[a,b]}f_n$$ exists and is the same for every such sequence. We call $a$ the value of the definite integral of $f$ over $[a,b]$, i.e. $$\int_a^bf:=a$$

The Riemann integral and the regulated integral of course agree on their common functions. Actually, the class of functions that are Riemann integrable is larger than the associated class of the regulated integral.

The reason why I went on such a odyssee to give a glimpse of this integral definition is that $f$ as you've defined it is a basis step function, where you even have provided a fitting finite partition:

As $f:[a,b]\to\mathbb{R}$ is $0$ except at the points $s_1,\dots,s_k$, we may define a fitting partition $P=\{a, s_1,\dots, s_k,b\}=\{x_0,\dots,x_n\}$, supposing that they are ordered ascendingly.

Now, as supposed by you, $f$ is constant on each open interval induced by $P$. Thus, the value of the integral of $f$ over $[a,b]$, as defined above, is just given as $$\int_a^bf=\sum_{j=1}^nc_j(x_j-x_{j-1})$$ for fitting constants $c_j$. Now, as $f$ is $0$ in all these intervals, the value of the constants is given by $c_j=0$ for all $j\in\{1,\dots,n\}$, i.e. $$\int_a^bf=\sum_{j=1}^n0\cdot(x_j-x_{j-1})=0$$

This is actually a important property of the regulated integral over step functions, as your function may obtain any value on the intersections of the partitioning intervals. But, as the partition is finite, you may only have finitely many of these.

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