text: # Interchange of summation and derivative
  
  I am looking at the Lemma in Ross's Stochastic Process textbook. The lemma says that
  
  $P\{S_{N(t)} \leq s \}=\bar{F}(t)+\int_0^s \bar{F}(t-y)\mathrm{d}m(y), t \geq s \geq 0$.
  
  In his proof, I am confused with the last step shown below:
  
  $=\bar{F} (t)+\sum_{n=1}^{\infty} \int_0^{\infty} P\{S_n \leq s, S_{n+1} > t |S_n = y\}\mathrm{d}F_n(y)$
  
  $= \bar{F}(t)+\sum_{n=1}^{\infty}\int_0^s \bar{F}(t-y)\mathrm{d}F_n(y)$
  
  $=\bar{F}(t) + \int_0^s \bar{F}(t-y)\mathrm{d}\left( \sum_{n=1}^{\infty} F_n(y)\right)$
  
  Although he mentions that the interchange of integral and summation is justified since all terms are non-negative in the bottom line, I believe it only accounts for why the summation symbol can go through the integral symbol.
  
  But I am really confused about the last step. Can someone explain the principle behind it and answer why such step is valid?
  
  For positive measures $\mu_1,\mu_2,\ldots$ it is indeed the case that $\mu:= \sum_n \mu_n$ is another measure and that $\int f d\mu = \sum_n \int f d\mu_n$ for suitable $f$ (e.g. positive, integrable). This is a measure theoretic fact and it is beyond the scope of Ross's book.
  The intuitive idea is that $d(\mu_1 + \cdots + \mu_n) = d\mu_1 + \cdots + d\mu_n$ obviously for $n$ finite, and then if the integrand $f$ is (say) positive, then these increase monotonically to the infinite sum.
  • For $f=1_A$ for a Borel set $A$, $\int f d\mu = \sum_n \int f d\mu_n$ just says $\mu(A) = \sum_n \mu_n(A)$ which is the definition. By linearity we can extend to $f=\sum a_n 1_{A_n}$ and then all positive measurable functions are increasing limits of such functions, so the monotone convergence theorem finishes the proof. – nullUser Feb 7 '15 at 6:35
explanation: The text extract is a detailed discussion of a mathematical concept from a stochastic processes textbook. It includes a specific lemma and a proof, with the user expressing confusion about the last step of the proof. The user asks for an explanation of the principle behind the interchange of the integral and summation in the last step of the proof.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Interchange of summation and derivative
  
  I am looking at the Lemma in Ross's Stochastic Process textbook. The lemma says that
  
  $P\{S_{N(t)} \leq s \}=\bar{F}(t)+\int_0^s \bar{F}(t-y)\mathrm{d}m(y), t \geq s \geq 0$.
  
  In his proof, I am confused with the last step shown below:
  
  $=\bar{F} (t)+\sum_{n=1}^{\infty} \int_0^{\infty} P\{S_n \leq s, S_{n+1} > t |S_n = y\}\mathrm{d}F_n(y)$
  
  $= \bar{F}(t)+\sum_{n=1}^{\infty}\int_0^s \bar{F}(t-y)\mathrm{d}F_n(y)$
  
  $=\bar{F}(t) + \int_0^s \bar{F}(t-y)\mathrm{d}\left( \sum_{n=1}^{\infty} F_n(y)\right)$
  
  Although he mentions that the interchange of integral and summation is justified since all terms are non-negative in the bottom line, I believe it only accounts for why the summation symbol can go through the integral symbol.
  
  But I am really confused about the last step. Can someone explain the principle behind it and answer why such step is valid?
  
  For positive measures $\mu_1,\mu_2,\ldots$ it is indeed the case that $\mu:= \sum_n \mu_n$ is another measure and that $\int f d\mu = \sum_n \int f d\mu_n$ for suitable $f$ (e.g. positive, integrable). This is a measure theoretic fact and it is beyond the scope of Ross's book.
  The intuitive idea is that $d(\mu_1 + \cdots + \mu_n) = d\mu_1 + \cdots + d\mu_n$ obviously for $n$ finite, and then if the integrand $f$ is (say) positive, then these increase monotonically to the infinite sum.
  • For $f=1_A$ for a Borel set $A$, $\int f d\mu = \sum_n \int f d\mu_n$ just says $\mu(A) = \sum_n \mu_n(A)$ which is the definition. By linearity we can extend to $f=\sum a_n 1_{A_n}$ and then all positive measurable functions are increasing limits of such functions, so the monotone convergence theorem finishes the proof. – nullUser Feb 7 '15 at 6:35
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a detailed discussion of a mathematical concept from a stochastic processes textbook. It includes a specific lemma and a proof, with the user expressing confusion about the last step of the proof. The user asks for an explanation of the principle behind the interchange of the integral and summation in the last step of the proof.
  
  Label: question-and-answer