# Problem in understanding the proof of Lebesgue-Radon-Nikodym Theorem

On Folland's Real Analysis book page $90$, the Lebesgue-Radon-Nikodym Theorem is given as

Let $\nu$ be a $\sigma$-finite signed measure and $\mu$ a $\sigma$-finite positive measure on $(X,\mathcal{M})$. There exists unique $\sigma$-finite signed measure $\lambda,\rho$ on $(X,\mathcal{M})$ such that $\lambda\perp \mu$, $\rho\ll\mu$, and $\nu=\lambda+\rho$. Moreover, there is an extended $\mu$-integrable function $f: X\to\mathbb{R}$ such that $d\rho=fd\mu$, and any two functions are equal $\mu$-a.e.

To prove this theorem, we first can consider the case that $\nu$ and $\mu$ are "finite" and "positive". Then, we can extend that to the case where $\nu$ and $\mu$ are "$\sigma-$finite" and "positive". Finally, since $\nu = \nu^+ - \nu^-$, we can conclude that for signed measure $\nu$.

But I have problem in understanding the second step. In this step, we can write $X = \cup_j A_j$ where $A_j$'s are disjoint and $\nu(A_j)< \infty$ and $\mu(A_j) < \infty$. Then, we can define, $\nu_j(E) = \nu(E \cap A_j)$ and $\mu_j(E) = \mu(E \cap A_j)$ where $\nu_j$ and $\mu_j$ are finite. So, from the results of the first step, we know that $\lambda_j\perp \mu_j$, $\rho_j\ll\mu_j$, and $\nu_j=\lambda_j+\rho_j$, $d\rho_j=f_jd\mu_j$. But then, it says that if we define $\lambda = \sum_j \lambda_j$ and $f = \sum_j f_j$, we have $\nu=\lambda+\rho$ where $d\rho = fd\mu$.

We know that $\rho_j\ll\mu_j$ and $\lambda_j\perp \mu_j$. To show that $\rho\ll\mu$ and $\lambda \perp \mu$, is it true to say that since for every $j$, $\rho_j\ll\mu_j$ and $\lambda_j\perp \mu_j$, then we can conclude that $\sum_j\rho_j\ll \sum_j\mu_j$ and $\sum_j\lambda_j\perp \sum_j\mu_j$?

• You've written out the proof of the second step, but which part do you not understand? Nov 27, 2015 at 1:57
• @angryavian I don't understand how we can reach $\nu = \lambda + \rho$ from $\nu_j = \lambda_j + \rho_j$? Nov 27, 2015 at 2:06

The sets $A_j$ are disjoint, and so $\nu_j\perp\nu_k$ for $j\ne k$. Then by countable subadditivity $$\sum_j\nu_j(E)=\sum_j\nu(E\cap A_j).$$ Again, the sets $E\cap A_j$ are disjoint, so $$=\nu\left(E\cap(\cup_jA_j)\right)=\nu(E\cap X)=\nu(E).$$

Similar results apply to $\lambda_j$ and $\rho_j=f_jd\mu_j$. You can use this to show that $\lambda$, $\rho$ are mutually singular unsigned measures, and that $\lambda\perp \mu$, $\rho\ll \mu$. Just write them out as the sum, remembering that the $A_j$ are disjoint.

Does that help? The proof that you have written is complete, what in particular about it doesn't make sense to you?

EDIT: In response to your edits, we will show that $\rho\ll \mu$, which is equivalent to showing $\rho=fd\mu$. By definition of $\rho$ $$\rho(E)=\sum_j\rho_j(E).$$ As $\rho_j=f_jd\mu$: $$=\sum_j\int_X\chi_Ef_jd\mu.$$ As the $f_j$ are positive we can use the Monotone Convergence Theorem to bring the sum inside the integral: $$=\int_X\chi_E\sum_jf_jd\mu=\int_X\chi_Efd\mu=\int_Efd\mu.$$ Thus $\rho=fd\mu$.

To show $\lambda\perp\mu$, suppose $E$ has $\mu$-measure zero, and define $E_i=E\cap A_i$. The result them follows from writing out $\mu=\sum_i\mu_i$, and noting that $\mu_i\perp\mu_j$ for $i\ne j$.

• Thanks for your response. I edited my question and specified the part I have problem with. Nov 27, 2015 at 2:19
• I have updated my answer. Nov 27, 2015 at 2:35
• Why is it that $\mu \perp \lambda$? I am not really able to see it without considering some ugly sets such as in my post. yesterday

The other answer to this post excellently explains with the Monotone Convergence Theorem why $$\rho<<\mu$$, but MCT is not really needed and there is a one line argument:

$$\mu(E)=0\Rightarrow \mu_j(E)=0\Rightarrow \rho_j(E)=0\Rightarrow \rho(E)=\sum\rho_j(E)=0$$

I believe we can also make it clearer why $$\lambda\perp \mu$$ and this to me is less immediate.

Let $$X=\dot{\cup} A_j$$ where each $$A_j$$ is $$\mu$$ and $$\lambda$$ finite. Folland defines $$\mu_j(E)=\mu(E\cap A_j)$$ and builds $$\lambda_j$$ such that $$\lambda_j\perp \mu_j$$ and $$\lambda_j$$ is zero in $$A_j^C$$. Observation I: In the notation above, $$\lambda_j \perp \mu$$.

We only know $$\lambda_j \perp \mu_j$$. Let $$X=N_j\dot{\cup} K_j$$ such that $$\lambda_j$$ is zero in $$K_j$$ and $$\mu_j$$ is zero in $$N_j$$. Let us define:

$$N_j \cap A_j \quad \text{and}\quad K_j \cup A_j^C$$

They are clearly disjoint and make up all of $$X$$. Furthermore, $$\mu$$ is zero in $$N_j\cap A_j$$ (because $$\mu$$ and $$\mu_j$$ coincide in $$A_j$$) and $$\lambda_j$$ is zero in $$K_j\cup A_j^C$$ but this is precisely what is meant by $$\lambda_j \perp \mu$$.

Observation II: If we have $$\lambda_j\perp \mu$$, we need $$\lambda=\sum \lambda_j \perp \mu$$.

Each $$\lambda_j$$ is zero in $$U_j$$ and $$\mu$$ is zero in $$U_j^C$$. Clearly: $$\mu(\cup U_j^C )\leq \sum \mu(U_j^C)=0$$ $$\lambda(\cap U_j)=\sum \lambda_j (\cap U_j)\leq \sum \lambda_j(U_j)=0$$

Thus $$\lambda \perp \mu$$, as we set out to prove.