Integration with respect to Dirac measure using a step function

Question

I have troubles to prove that $\int f(x) d\delta_x(A)=f(x)$ using a step function like $g(x)=\sum_i c_i \mathbb{1_{A_i}}(x) $.

I have tried this :

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

And as we know in general that $\int_A dP(x)= P(A)$ we get something like this (This is where I get confused):

$$\int g(x)d\delta_x(A)=\sum_i c_i\delta_ \mathbb{{A_i}}(A)$$

which doesn't lead to the result $\int g(x)d\delta_x(A)=g(x)$

Any comments? Help? Thank you

Answer 1

Edit to make things more clear.

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Let $x$ be fixed.

Then $\delta_x$ denotes a measure wich is prescribed by $A\mapsto 1_A(x)$ on measurable sets $A$.

So by definition $\delta_x(A)=1_A(x)$ for measurable sets $A$.

For every measure $\mu$ and every measurable set $A$ the following notations are used to denote an integral over a characteristic function:

• $\int1_Ad\mu=\mu(A)$
• $\int_Ad\mu=\mu(A)$
• $\int1_A(y)d\mu(y)=\mu(A)$

Applying that on $\delta_x$ we get

• $\int1_Ad\delta_x=\delta_x(A)$
• $\int_Ad\delta_x=\delta_x(A)$
• $\int1_A(y)d\delta_x(y)=\delta_x(A)$

For step function $g$ prescribed by $y\mapsto\sum_{i=1}^{n}c_{i}1_{A_{i}}\left(y\right)$ we find:

$$\int g\left(y\right)d\delta_{x}\left(y\right)=\sum_{i=1}^{n}c_{i}\int1_{A_{i}}\left(y\right)d\delta_{x}\left(y\right)=\sum_{i=1}^{n}c_{i}\delta_{x}\left(A_{i}\right)=\sum_{i=1}^{n}c_{i}1_{A_{i}}\left(x\right)=g\left(x\right)$$

Answer 2

You want to show $\int_X f d \delta_x=f(x)$. First consider a simple function $s=\sum_i c_i 1_{A_i}$ where the $A_i$ are disjoint. For convenience we can assume WLOG that $\bigcup A_i=X$ (by taking a $c_i$ to be zero if need be). Then there is exactly one $i$ with $x \in A_i$, call this $i(x)$. Then $\int_X s d \delta_x = c_{i(x)} \delta_x(A_{i(x)})=c_{i(x)}=s(x)$. Now extend in the usual way.