I am self-studying measure theory and I ran into this problem when studying $L_p$ space.
Assume we have a measurable space $(X,\Sigma, \mu)$ where $\Sigma$ is a $\sigma$-algebra on $X$, and $\mu$ is a measure. Let $L$ be the set of all pairs $(Y,f)$ where $Y \in \Sigma$ and such that $\mu(X-Y)=0$. Let $f: Y\rightarrow C$ be a measurable and integrable function with respect to Y, i.e, $$\int_Y |f| d\mu < +\infty.$$
Now, let $L_1$ be the quotient set $L/\sim$ where the equivalence relation $\sim$ is given as
$$(Y,f) \sim (Y', f')\ \text{if}\ f(x) = f'(x)\ \text{for all $x$ in } Y \cap Y'$$
Show that "$\sim$" defined above is indeed a equivalence relation.
Showing $a\sim a$ and $a\sim b \rightarrow b\sim a$ is trivial, but in order to show $a\sim b$, $b\sim c$, implies $a\sim c$ requires us to show that the following statement
$$\text{Assume we have a set of Y's},\ \{Y_1, Y_2, \dots\}.\ \text{If x is in}\ Y_i \cap Y_j\ \text{for some}\ i\neq j,\ \text{then we would have}\ \ x\in Y_i\ \forall i\in \mathbb{N} $$
Anyone has any thoughts how to show the statement above?
It is simply not true. let $X=\{1,2,3,4\}$, $\Sigma=2^X$, and $\mu$ is defined so that $\mu(A)=0$ if $1\notin A$ and $\mu(A)=1$ if $1\in A$. Given this setting $1\in Y$, then $\mu(X-Y)=0$.
Let $Y_1=\{1,2,3\}$, $Y_2=\{1,2,4\}$, $Y_3=\{1,3,4\}$ and define $f_1,f_2,f_3$ via (hopefully, the notation is clear) $f_1=(1:1,2:2,3:3,4:4)$, $f_2=(1:1,2:2,3:3,4:4)$, and $f_3=(1:1,2:2,3:1,4:4)$. It is easily verified that $(Y_1,f_1)\sim(Y_2,f_2)$ and $(Y_2,f_2)\sim (Y_3,f_3)$, but not $(Y_1,f_1)\sim(Y_3,f_3).$
You get an actual equivalence relation if you take $~$ to be given by $(Y,f)\sim (Y',f')$ if $\mu\{x\in X:f(x)\neq f'(x)\}=0$, or, equivalently, by $(Y,f)\sim (Y',f')$ if $\mu\{x\in Y_1\cap Y_2:f(x)\neq f'(x)\}=0$.