A fallacious proof of the existece of non-Borel sets.

Question

Let $\lambda$ be Lebesgue-Borel measure and suppose that $\lambda(X)$ is defined for every $X\subset[0,1]$, in which case $\lambda(X) \in [0,1]$ by monotonicity. Define the set $B$ by

$$B=\{\lambda(X):X\subset[0,1], \lambda(X)\not\in X\}$$

It is a subset of $[0,1]$. Upon testing $\lambda(B)$ for membership in $B$ we find that the statements $\lambda(B)\in B$ and $\lambda(B)\not \in B$ are equivalent, a contradiction. Hence some $X\subset[0,1]$ is not Borel.

What is the problem in this Russell's paradox flavoured proof?

In my opinion the proof merely shows that, under the the hypothesis that every subset of $[0,1]$ is measurable, such set $B$ cannot exist, not that the initial hypothesis is false.

EDIT: $\lambda:\mathcal{B}(\mathbb{R})\to [0,\infty]$ is assumed to be the Lebesgue-Borel measure, with all its known properties. The question is whether $\mathcal{B}(\mathbb{R})=\mathcal{P}(\mathbb{R})$.

Answer 1

$$\lambda(B)\in B\tag{1}$$ just means that there is some $C$ such that $\lambda(C)=\lambda(B)$ and $\lambda(C)\notin C.$

For example, if singletons all have measure $0,$ then every $\lambda(X)\in B.$ Because you can use $C=X \setminus\{\lambda(X)\}.$

So there is no contradiction in $(1).$

Answer 2

This isn't even justified:

Upon testing $\lambda(B)$ for membership in $B$ we find that the statements $\lambda(B)\in B$ and $\lambda(B)\not \in B$ are equivalent...

How?

Note that you only assume that $\lambda$ is a monotonic function, not that it's the real Lebesgue-Borel measure. Okay, let us define $\lambda(X)=0$ for all $X$. Then for $X=\{1\}$, we have $\lambda(X)\not\in X$, so $\lambda(X)=0\in B$. In fact, $B=\{0\}$. And it turns out that $\lambda(B)\in B$.