Lebesgue measure on uncountable union of singletons I have the following excerise:

Let $\lambda$ be the Lebesgue measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. What is the wrong with the following arguement?
$$\lambda([0,1])=\lambda(\bigcup_{x\in [0,1]}\{x\})=\sum_{x\in [0,1]}\lambda(\{x\})=\sum_{x\in [0,1]}0=0$$


My attempt:
It's known that $\lambda(A)=0, \forall $ countably subsets of $\mathbb{R}.$
Further is it known that $(0,1)$ is uncountable which implies that $(0,1)\cup\{0,1\}=[0,1]$ is uncountable.
My arguement is that sense $[0,1]$ is uncountable, $\lambda([0,1])\neq 0.$
Am i on the right track?
 A: Well, yes you are; $[0, 1]$ is not countable. When you know that, then why do you conclude
$$
\lambda \left( \bigcup_{x \in [0, 1]} \lbrace x \rbrace \right) = \sum_{x \in [0, 1]} \lambda\big( \lbrace x \rbrace\big) \quad?
$$
A: uncountability does not imply that $\lambda(S)\neq 0$ when $S$ is an uncountable set.
the best example is Cantor's set $C$ which is uncountable and has measure 0.
In the case of $[0,1]$ since it is not countable then you can not jump from union to sum.
$\lambda(\cup_{x\in [0 ,1]}\{x\})\neq \sum{\lambda \{x\}}$
$\lambda$ is $\sigma-$ additive which means that for a countable union of disjoint sets:
$\lambda (\cup_n{A_n})=\sum_n{\lambda(A_n)}$
A: There's 1 issue with the ff equality that for some reason is not covered in the other answers
$$\lambda(\bigcup_{x\in [0,1]}\{x\})=\sum_{x\in [0,1]}\lambda(\{x\})$$
Yes, Lebesgue measure $\lambda$ and actually measure in general is just countably / $\sigma$-additive, not necessarily 'uncountably additive'. But more importantly, what would 'uncountable additivity' even look like?
The other answers seem to be saying just that the above equality is wrong when actually I think it's vacuously correct in that the RHS is nonsensical or at least not well-defined.
When you use this term $\sum$ you're 'summing' over at most a countable set. The very reason why integration exists, afaik, is to 'sum' over an uncountable set. I mean
$$\int_{[7,8]} 4 dx = 4 \times 1 = 4$$
like the rectangle $[7,8] \times [0,4]$ has height $4$ and base $1$, but this rectangle is composed of (uncountably infinite) 'rectangles' of zero area namely $\{\{x\} \times [0,4]\}_{x \in [7,8]}$ in the sense that their heights are still 4 but the bases have 'length' or 'measure' zero namely the 'length' or 'measure' of singletons $\{x\}_{x \in [7,8]}$.
Similarly, rectangles have 'measure' zero in 3D space. Yet the 'sum' of all these objects of 'measure' zero are cubes which have positive measure.
So uncountably adding things will lead to $0=4$ because assuming this kind of operation '$\sum_{x \in [7,8]}$' is well-defined:
$$\sum_{x \in [7,8]} \text{'length'}(\{x\}) \times 4$$
$$= \sum_{x \in [7,8]} 0 \times 4 $$
$$= \sum_{x \in [7,8]} 0 = 0$$
meanwhile
$$\sum_{x \in [7,8]} \text{'length'}(\{x\}) \times 4 = \int_{[7,8]} 4 dx = 4 \times 1 = 4$$
Even when you do those non-measurable things like Vitali sets for Lebesgue measure the summations are still over countable sets:
From Royden and Fitzpatrick:



From Wikipedia:



