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 $\sum_j\lambda_j\perp \sum_j\mu_j$?

  • $\begingroup$ You've written out the proof of the second step, but which part do you not understand? $\endgroup$
    – angryavian
    Nov 27, 2015 at 1:57
  • $\begingroup$ @angryavian I don't understand how we can reach $\nu = \lambda + \rho$ from $\nu_j = \lambda_j + \rho_j$? $\endgroup$
    – m0_as
    Nov 27, 2015 at 2:06

2 Answers 2


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 $i\ne j$.

  • $\begingroup$ Thanks for your response. I edited my question and specified the part I have problem with. $\endgroup$
    – m0_as
    Nov 27, 2015 at 2:19
  • $\begingroup$ I have updated my answer. $\endgroup$ Nov 27, 2015 at 2:35
  • $\begingroup$ 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. $\endgroup$
    – Kadmos

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 $A_j^C$.

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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.


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