How to integrate a function that equals 1 at one specific point?

Question

I am having trouble integrating a function that looks something like the following:

$$\int\limits_0^t I(X=u)\;du$$ where $I$ is an indicator function that equals 1 when $X=u$ and $0$ otherwise.

My intuition is telling me that it is equal to $I(X\leq t)$ but I am unsure how to prove this.

Are you really integrating w.r.t. $u$ and using $u$ as the indicator function target value? — apnorton, Sep 03 '13 at 03:18
so $I(X=u)=1$ only for $u=X$? $X$ is not changing? then the set ${X}$ is null — obataku, Sep 03 '13 at 03:19
Anorton: yes. Oldrinb: $I(X=u)=1$ only for $u=X$ but $u$ is what I am integrating with respect to. — NewNameStat, Sep 03 '13 at 03:23
Is this the Riemann integral or the Lebesgue integral or something simpler? — Ben Millwood, Sep 03 '13 at 03:24
I hope my detailed explanation below, from first principles, convinces you. — MJD, Sep 03 '13 at 04:06
MJD's answer is correct. You might be looking for the Dirac delta "function" (technically a distribution which puts your question beyond Reimann integration). That has the "indicator" behavior you suggested, i.e. $\int^y_x \delta(u) du = 1$ iff $0 \in (x,y)$ and $= 0$ otherwise. But as I said, it is not a well defined function in the proper sense, since it is infinite at $u=0$. — user1795638, Sep 03 '13 at 04:19

MJD · Accepted Answer · 2013-09-03T04:12:45.777

tl;dr you can modify a function any way you want at any finite number of points without affecting its Riemann integral in the slightest. Since your function $I$ is a constant zero function, modified at one point, it has the same Riemann integral as the constant zero function.

However, in hideous detail…

The definition of the Riemann integral $$\int_a^b f(x)\; dx$$ is as follows: we consider a partition $\Pi$ of the interval $[a, b]$ into subintervals $a = x_0 < x_1 < \cdots < x_n = b$, and then calculate the upper sum $$M(\Pi,f) = \sum_{i=0}^{n-1} (x_{i+1}-x_i)\sup_{x_i\le x\le x_{i+1}} f(x)$$ and the lower sum $$m(\Pi, f) = \sum_{i=0}^{n-1} (x_{i+1}-x_i)\inf_{x_i\le x\le x_{i+1}} f(x)$$ for each such partition.

The "upper integral" is defined to be the infimum (greatest lower bound) of these upper sums: $$ \overline{\int_a^b} f(x)\; dx = \inf M(\Pi, f)$$ and the "lower integral" similarly is the supremum (least upper bound) of the lower sums: $$\underline{\int_a^b} f(x)\;dx = \sup m(\Pi, f),$$

If the upper and lower integrals are equal, we define the Riemann integral to be equal to their common value: $$ \int_a^b f(x)\; dx = \overline{\int_a^b} f(x)\;dx = \underline{\int_a^b} f(x)\;dx.$$ This is all straight from Rudin Principles of Mathematical Analysis pages 120–121.

Now let's consider your function $I$ which is zero everywhere except at $x=k$. Evidently the parts of $\Pi$ that don't contain $k$ will contribute zero to the sums $m(\Pi, I)$ and $M(\Pi, I)$ since $\sup I = \inf I = 0$ on these intervals. So we need only worry about the one interval in each partition, say $[x_p, x_{p+1}]$, that contains $k$.

Even on this one special interval, $\inf I$ is still zero, so $m(\Pi, I) = 0$ for every partition $\Pi$ and therefore the "lower integral" $$\underline{\int_a^b} I(x)\;dx = 0.$$ So if the Riemann integral exists for your function $I$, it must be zero.

Fortunately, the "upper integral" is also zero, because the part of $\Pi$ in which $k$ lies can have $x_{p+1} - x_p$ arbitrarily small, say $\epsilon$, and therefore $$\begin{align} M(\Pi,I) & = \sum_{i=0}^{n-1} (x_{i+1}-x_i)\sup_{x_i\le x\le x_{i+1}} I(x)\\ & = (x_{p+1} - x_p)\sup_{x_p\le x\le x_{p+1}} I(x)\\ & = (x_{p+1} - x_p)I(k)\\ & = \epsilon \cdot 1 \\ & = \epsilon \end{align}$$

and by choosing $\Pi$ suitably we can make $\epsilon$ as close as we like to zero, and therefore the "upper integral" $$\overline{\int_a^b} I(x)\;dx = \inf M(\Pi, I) $$ is zero.

So the upper and lower integrals are both zero, and therefore (by definition) so is the Riemann integral.

How to integrate a function that equals 1 at one specific point?

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