Let there be a measurable space , i.e. a basic set together with a σ-algebra defined on it . An associated mapping is defined for each point , which assigns the value to each set if it contains and the value if it does not contain:
${\ displaystyle (\ Omega, {\ mathcal {A}})}$${\ displaystyle \ Omega}$${\ displaystyle {\ mathcal {A}}}$${\ displaystyle z \ in \ Omega}$${\ displaystyle \ delta _ {z}}$${\ displaystyle A \ in {\ mathcal {A}}}$${\ displaystyle 1}$${\ displaystyle z}$${\ displaystyle 0}$${\ displaystyle z}$

${\ displaystyle \ delta _ {z} (A): = {\ begin {cases} 1 \, & {\ text {if}} z \ in A \, \\ 0 \, & \ mathrm {otherwise} \. \ end {cases}}}$

The mapping is then a measure and is called a Dirac measure or point measure in a point . Because of there is even a probability measure and a probability space . This can be used to define the Dirac distribution . In the Dirac dimension , the unit mass is concentrated in the point . It follows that the measure is finite, in particular the measure space is σ-finite .
${\ displaystyle \ delta _ {z} \ colon {\ mathcal {A}} \ to [0,1]}$${\ displaystyle z}$${\ displaystyle \ delta _ {z} (\ Omega) = 1}$${\ displaystyle \ delta _ {z}}$${\ displaystyle (\ Omega, {\ mathcal {A}}, \ delta _ {z})}$${\ displaystyle \ delta _ {z}}$${\ displaystyle z}$

With the help of the characteristic function one can also use the defining equation
${\ displaystyle \ chi}$

for all and express.
${\ displaystyle z \ in \ Omega}$${\ displaystyle A \ in {\ mathcal {A}}}$

Dirac integral

The Dirac integral of the function is defined as the Lebesgue integral under the Dirac measure. Instead of the Lebesgue measure, the Dirac measure is used to calculate the integral. This gives f for the integral of any function.
${\ displaystyle f \ colon A \ to \ mathbb {R}}$

${\ displaystyle \ int _ {A} f \ mathrm {d} \ delta _ {z} = {\ begin {cases} f (z) & z \ in A \\ 0 & z \ not \ in A \ end {cases}} }$

Reason

The mapping is a non-negative measurable function . The Lebesgue integral of the function under the Dirac measure is defined as follows.
${\ displaystyle f \ colon A \ to \ mathbb {R}}$

${\ displaystyle f_ {n}}$is an arbitrary sequence of simple functions which converge pointwise and monotonically increasing to . A simple function is a non-negative measurable function that only takes a finite number of function values. be the number of function values ; be the (measurable) quantities on which the function assumes the value . The integral of a simple function is thus defined as follows:
${\ displaystyle f}$${\ displaystyle \ alpha _ {i}}$${\ displaystyle m}$${\ displaystyle \ alpha _ {i}}$${\ displaystyle A_ {i}}$${\ displaystyle f_ {n}}$${\ displaystyle \ alpha _ {i}}$

Is , then is certainly not an element of any of the subsets . Then the Dirac measure of all is also zero. As a result, the integral over all is zero.
${\ displaystyle z \ notin A}$${\ displaystyle z}$${\ displaystyle A_ {i}}$${\ displaystyle A_ {i}}$${\ displaystyle A}$

If for any one, the Dirac measure of is equal ; the Dirac measure for all other sets is then zero. For the integral of the simple functions we get:
${\ displaystyle z \ in A_ {j} (n)}$${\ displaystyle j}$${\ displaystyle A_ {j} (n)}$${\ displaystyle 1}$${\ displaystyle A_ {i} (n)}$${\ displaystyle f_ {n}}$

So the Dirac integral is equal to the function value at the point if is.
${\ displaystyle z}$${\ displaystyle z \ in A}$

Another line of evidence is as follows:

For everyone and applies
${\ displaystyle z \ in \ Omega}$${\ displaystyle A \ in {\ mathcal {A}}}$

${\ displaystyle {\ begin {aligned} \ int \ limits _ {A} f \, \ mathrm {d} \ delta _ {z} & = \ int \ limits _ {A \ cap f ^ {- 1} (\ {f (z) \})} f \, \ mathrm {d} \ delta _ {z} + \ int \ limits _ {A \ setminus f ^ {- 1} (\ {f (z) \})} f \, \ mathrm {d} \ delta _ {z} \\ & = \ int \ limits _ {\ {x \ in A \ mid f (x) = f (z) \}} f \, \ mathrm { d} \ delta _ {z} + \ int \ limits _ {\ {x \ in A \ mid f (x) \ neq f (z) \}} f \, \ mathrm {d} \ delta _ {z} \\ & = f (z) \ delta _ {z} (A) +0 \\ & = f (z) \ delta _ {z} (A) \\ & = {\ begin {cases} f (z) & z \ in A \\ 0 & z \ not \ in A \ end {cases}} \ end {aligned}}}$

As a one-element subset of is . Archetypes of measurable quantities are measurable. So is and accordingly also the quantities that are integrated above.
${\ displaystyle \ mathbb {R}}$${\ displaystyle \ {f (z) \} \ in {\ mathcal {B}}}$${\ displaystyle f ^ {- 1} (\ {f (z) \}) \ in {\ mathcal {A}}}$

If so, integration via and is also possible.
${\ displaystyle \ {z \} \ in {\ mathcal {A}}}$${\ displaystyle A \ cap \ {z \}}$${\ displaystyle A \ setminus \ {z \}}$