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Conventions

[tex]X[/tex] is a set.

[tex]\mathcal{A}[/tex] is a [itex]\sigma[/itex]-algebra.

Suppose that I have a measure space [itex](X,\mathcal{A},\mu)[/itex] and an [itex]\mathcal{A}[/itex]-measurable function:

[tex]f\,:\,X\rightarrow[0,\infty][/tex]

All pretty regular stuff. Now, I have a "supposed" measure defined as

[tex]\nu(E):=\int_E f\mbox{d}\mu[/tex]

for [itex]E\in \mathcal{A}[/itex]. My endeavor is to understand whether or not this function actually defines a measure on [itex]\mathcal{A}[/itex] and furthermore, understand exactly what is needed to do to work out if a given function is a measure.

The first thing I noticed was that this function is defined using a Lebesgue integral. I think it is a Lebesgue integral because I am integrating the function over a set E. Since E is in A then I know E is closed under complements and unions. Secondly the integral is with respect to [itex]\mu[/itex], another measure different to [itex]\nu[/itex] (which Im trying to work out if it is a measure).

So if [itex]\nu[/itex] is going to be a measure then it has to satisfy some properties:

[tex]1) \quad \nu(\emptyset) = 0[/tex]

and

[tex]2) \quad \nu\left(\bigcup_{i=1}^{\infty}E_i\right) = \sum_{i=1}^{\infty}\nu(E_i)[/tex]

So the first thing I did was see what happens to the empty set. Let me know if this looks right

[tex]\nu(\emptyset) = \int_{\emptyset}f\mbox{d}\mu[/tex]

[tex]= \lim_{i\rightarrow\infty}\sum_{i=1}^{\infty}f_i\mu(E)[/tex] where [tex]\lim_{i\rightarrow\infty}f_i = f[/tex]

[tex]= \lim_{i\rightarrow\infty}\sum_{i=1}^{\infty}f_i\cdot 0[/tex]

[tex]=0[/tex]

Basically I was looking for a way to get "[itex]\mu(\emptyset)[/itex]" into the picture because I know that the measure of the emptyset is zero - always. So I converted the Lebesgue integral into an equation involving the limit of the sums of a sequence of [itex]f_i[/itex]'s - which involves taking the measure of the set. Im not sure if this is the right way to do it but it is the only way that I could see that I would get zero into the equation.

[tex]X[/tex] is a set.

[tex]\mathcal{A}[/tex] is a [itex]\sigma[/itex]-algebra.

Suppose that I have a measure space [itex](X,\mathcal{A},\mu)[/itex] and an [itex]\mathcal{A}[/itex]-measurable function:

[tex]f\,:\,X\rightarrow[0,\infty][/tex]

All pretty regular stuff. Now, I have a "supposed" measure defined as

[tex]\nu(E):=\int_E f\mbox{d}\mu[/tex]

for [itex]E\in \mathcal{A}[/itex]. My endeavor is to understand whether or not this function actually defines a measure on [itex]\mathcal{A}[/itex] and furthermore, understand exactly what is needed to do to work out if a given function is a measure.

The first thing I noticed was that this function is defined using a Lebesgue integral. I think it is a Lebesgue integral because I am integrating the function over a set E. Since E is in A then I know E is closed under complements and unions. Secondly the integral is with respect to [itex]\mu[/itex], another measure different to [itex]\nu[/itex] (which Im trying to work out if it is a measure).

So if [itex]\nu[/itex] is going to be a measure then it has to satisfy some properties:

[tex]1) \quad \nu(\emptyset) = 0[/tex]

and

[tex]2) \quad \nu\left(\bigcup_{i=1}^{\infty}E_i\right) = \sum_{i=1}^{\infty}\nu(E_i)[/tex]

So the first thing I did was see what happens to the empty set. Let me know if this looks right

[tex]\nu(\emptyset) = \int_{\emptyset}f\mbox{d}\mu[/tex]

[tex]= \lim_{i\rightarrow\infty}\sum_{i=1}^{\infty}f_i\mu(E)[/tex] where [tex]\lim_{i\rightarrow\infty}f_i = f[/tex]

[tex]= \lim_{i\rightarrow\infty}\sum_{i=1}^{\infty}f_i\cdot 0[/tex]

[tex]=0[/tex]

Basically I was looking for a way to get "[itex]\mu(\emptyset)[/itex]" into the picture because I know that the measure of the emptyset is zero - always. So I converted the Lebesgue integral into an equation involving the limit of the sums of a sequence of [itex]f_i[/itex]'s - which involves taking the measure of the set. Im not sure if this is the right way to do it but it is the only way that I could see that I would get zero into the equation.

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