# Integration with respect to Dirac measure

The Dirac measure is defined by $$\delta_x(A)= \begin{cases} 1 &\text{if x \in A}\\ 0 &\text{if x \notin A}\\ \end{cases}$$ Let $f:X\rightarrow \mathbb{R}$ be a function. Can anyone show me why $\int f \, d\delta_x=f(x)$? Any help is appreciated.

Hint If $g=\sum_i c_i 1_{A_i}$ is a step function, then

$$\int g d \delta_x = \sum_i c_i \delta_x(A_i)=\sum_i c_i 1_{A_i}(x)=g(x)$$

• The function given is $f:X\rightarrow \mathbb{R}$. Why are u using $g=\sum_i c_i 1_{A_i}$? Mar 27, 2013 at 14:50
• @NZhang Because the definition of integrals with respect with measures is typically defined via those functions. You first define the integral for step functions (i.e. linear sum of characteristic functions) then extended it to positive functions and then to all measurable functions... So as soon as you can prove a formula like yours for step functions you get it for free in general..... Mar 27, 2013 at 14:57
• What's the difference between step function and simple function? Mar 27, 2013 at 15:20
• @NZhang They mean the same thing... I am used to call them step functions :) Mar 27, 2013 at 17:10
• Step functions are usually implemented using indicator of intervals, while simple functions are implemented using indicator of arbitrary measurable sets Oct 16, 2015 at 5:55

Let $g$ denote the constant function being equal to $f(x)$. Then $f=g$ almost surely with respect to $\delta_x$, i.e. the function $f$ is almost surely equal to the constant function taking on the value $f(x)$. This is easily seen since $$\delta_x(\{g=f\})=\delta_x(\{y\in X\mid g(y)=f(y)\})=1$$ since $x\in \{g=f\}$ is in that set. Now use that two functions who are identical almost surely have the same integral. That is $$f=g\quad\delta_x\text{-a.s}\;\Longrightarrow\;\int f\,\mathrm d\delta_x=\int g\,\mathrm d\delta_x=f(x)\int 1\,\mathrm d\delta_x=f(x)\cdot\delta_x(X)=f(x).$$

If this is done for the case where $f$ is everywhere nonnegative, you can probably figure out how to do the rest.

Partition the whole space $X$ into two sets: $\{x\}$ and the complement of $\{x\}$. Look at the simple function $$g(w)=\begin{cases} f(x) & \text{if }w=x, \\ 0 & \text{if }w\ne x. \end{cases}$$ Then $g\le f$ everywhere and $\int_X g \, d\delta_x=f(x)$. Therefore $\int_X f\,d\delta_x\le f(x)$.

If $h$ is any other simple function $\le f$ then one of the measurable sets on which $h$ is constant contains $x$, and so on that set the value of $h$ cannot exceed $f(x)$. The measure of that set is $1$, and the measure of each of the other measurable sets on which $h$ is constant is $0$. Hence $\int_X h\,d\delta_x\le f(x)$.

• The integral $\int_X g \,\mathrm d\delta_x$ may not be defined. You have implicitly assumed $\{x\}$ to be measurable. Mar 27, 2013 at 15:06
• That's true: All of this works only if $\{x\}$ is measurable. Mar 27, 2013 at 15:12