Can you give me an example of $A,B,C \subset{\mathbb{R}}$ with $A = B\setminus C$ but $\mu(A) \neq \mu(B) - \mu(C)$? [closed]

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.

• This is true for finite measures. Note that you have bot correctlt identified B and C in your example. Nov 23 '21 at 6:05
• @CrackedBauxite Oh thanks, fixed that just now by swapping the $t_1, t_2$ Nov 23 '21 at 6:09
• @CrackedBauxite What does the sentence "you have bot correctlt identified B and C" mean?
– 5xum
Nov 23 '21 at 6:38
• Generally one avoids subtraction with measures if the possibility of $\infty$ appears. Nov 23 '21 at 17:53

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 $$\mu(A) =\mu(B) -\mu(C)$$

• 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.
– S. G
Nov 23 '21 at 7:42
• +1 No idea why the downvote. Nov 23 '21 at 17:52
• Sir, is my answer correct ? Is there any mistake?
– S. G
Nov 23 '21 at 17:56
• Looks fine to me. There are always random downvotes out there. I find them particularly annoying when they do not indicate the reason. Nov 23 '21 at 18:18

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

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

• I'm guessing OP is assuming implicitly that $C \subset B$
– D_S
Nov 23 '21 at 6:41
• @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.
– 5xum
Nov 23 '21 at 8:00

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.