MHB Proving Absolute Convergence of a Real-Valued Function on a Sigma Algebra

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Let R be a sigma algebra and let $f$ be a real value function on R such that for a sequence
($A_{n}$) of disjoint members of R, we have that the sum of $f$($A_{n}$) over all n is equal to the image of the countable union under $f$. Prove that the sum of $f$($A_{n}$) is in fact absolutely convergent.
 
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Since $f$ is a real-valued function, it must be bounded. That is, there exists a constant M such that |$f$($A_{n}$)| $\leq$ M for all n. By the Principle of Finite Sums, we know that the sum of a sequence of finitely many non-negative real numbers is convergent if and only if the sequence of partial sums is increasing and bounded above. Since the sum of $f$($A_{n}$) over all n is equal to the image of the countable union under $f$, the sequence of partial sums is increasing and bounded above by M. Therefore, the sum of $f$($A_{n}$) is absolutely convergent.
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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