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I'm given the measure of $$ \mu(\{x: f(x) > t\}) $$ for all $t$. And in solving a problem, I said that
$$ \mu(\{x: t_1 < f(x) < t_2\}) = \mu(\{x: f(x) > t_1 \}) - \mu(\{x: f(x) > t_2 \}) $$ But, looking back and thinking about it, I'm unsure if this is even true, despite it being intuitive.

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  • $\begingroup$ This is true for finite measures. Note that you have bot correctlt identified B and C in your example. $\endgroup$ Nov 23 '21 at 6:05
  • $\begingroup$ @CrackedBauxite Oh thanks, fixed that just now by swapping the $t_1, t_2$ $\endgroup$
    – Obamafish
    Nov 23 '21 at 6:09
  • $\begingroup$ @CrackedBauxite What does the sentence "you have bot correctlt identified B and C" mean? $\endgroup$
    – 5xum
    Nov 23 '21 at 6:38
  • $\begingroup$ Generally one avoids subtraction with measures if the possibility of $\infty$ appears. $\endgroup$
    – copper.hat
    Nov 23 '21 at 17:53
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Let, $(X,{\scr{B}}(X),\mu)$ be a measure space.

$ B $ and $ C$ are measurable sets with $C\subset B$ and $\mu(C) <\infty$ , then your conclusion, $\mu(A) =\mu(B\setminus C) =\mu(B) - \mu(C) $ is true.

Because, \begin{align} \quad \mu(B) &=\mu(B \cap C) + \mu(B \cap C^c) \\ &= \mu(C) + \mu(B \setminus C) \end{align}

And, hence $\mu(B\setminus C) =\mu(B) - \mu(C) $$(\mu(C) <\infty ) $

Consider,$(\mathbb{R},{\scr{L}}(\mathbb{R}), m) $

$ (-\infty, 0) =\mathbb{R} \setminus [0, \infty) $

(all are Borel sets and so Lebesgue measurable)

And applying above property(excluding the possibility of finite measure of $C$)

$m\{(-\infty,0)\} =m\{\mathbb{R}\} - m\{[0,\infty ) \}$

Hence, $\infty=\infty -\infty$(!)

Hence, $A=B\setminus C$ doesn't imply $\mu(A) =\mu(B) -\mu(C) $

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  • $\begingroup$ Please leave a comment before going to Downvote. It will help me to find my mistake. So that i can make sure about my mistake and it will be really helpful for me as an average student. $\endgroup$
    – S. G
    Nov 23 '21 at 7:42
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    $\begingroup$ +1 No idea why the downvote. $\endgroup$
    – copper.hat
    Nov 23 '21 at 17:52
  • $\begingroup$ Sir, is my answer correct ? Is there any mistake? $\endgroup$
    – S. G
    Nov 23 '21 at 17:56
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    $\begingroup$ Looks fine to me. There are always random downvotes out there. I find them particularly annoying when they do not indicate the reason. $\endgroup$
    – copper.hat
    Nov 23 '21 at 18:18
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No. Take $B=\emptyset$ and let $C$ be any set with positive measure. Then, $A=\emptyset$, and the equality does not hold, since

$$\mu(A)=0\neq -\mu(C)=\mu(B)-\mu(C).$$

For example, if looking at $\mathbb R$ with the Borel measure, then you can take $A=B=\emptyset, C=[0,1]$, and you have

$$\mu(A)=0\neq-1=\mu(B)-\mu(C)$$

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    $\begingroup$ I'm guessing OP is assuming implicitly that $C \subset B$ $\endgroup$
    – D_S
    Nov 23 '21 at 6:41
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    $\begingroup$ @D_S I tend to answer questions the OP asks, I am not good at guessing what they wanted to say. If OP changes the question, I will change the answer. $\endgroup$
    – 5xum
    Nov 23 '21 at 8:00
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No. In general $$ B \cup C = C \cup (B \backslash C)$$ $$ \mu (A)= \mu(B \cup C) - \mu(C) $$

When $C \subset B$ it reduces to your expression.

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