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Consider a uniform random variable $U$ in the interval $(0,1)$. Trivially we have that

$$mathbbE[X|X<tfrac12]=frac14.$$

Now, based on my intuition, I would like to say that $(X|X<tfrac12)$ conditioned on the random variable $X,$ is a uniform random variable in $(0,tfrac12).$

How do we formally write and prove it, in terms of the "conditional expectation" formalism?

Another point of view on the same problem is the following: consider the following expression for two dependent uniform random variables $U,V$ in $1,2,dots,n$:

$$mathbbP(U<V)=sum_mmathbbP(U<m|V=m)mathbbP(V=m)$$

If now I consider the conditional probability $mathbbP(U<V|U),$ I would like to generalize the previous expression to:

$$mathbbP(U<V|U)=sum_mmathbbP(U<m|V=m,U)mathbbP(V=m|U)$$

but I do not know how to exactly define the expression $mathbbP(U<m|V=m,U).$ Any help?

The conditional distribution of $X$ given the event $X < 1/2$ is indeed uniform on $[0, 1/2]$.

One way is to show that for $t in [0, 1/2]$,
$$P(X le t mid X < 1/2) = fracP(X le t, X < 1/2)P(X < 1/2) = 2 P(X le t) = 2t,$$
which is the CDF of the uniform distribution on $[0, 1/2]$.

Unfortunately, I do not understand what you mean by "$(X mid X < 1/2)$ conditioned on the random variable $X$." Once you condition on $X$, then $X$ is no longer random. Given that you were looking for something with the uniform distribution on $[0, 1/2]$, I suspect you wanted the above.

by definition, if $P(A)>0$ , $E(X|A)=fracE(XI_A)p(A)$ $hspace.5cm$ (1)

by your note "in terms of the "conditional expectation" formalism?"

you point that conditional probability defined based on conditional expectation.

$P(Xleq a| X>3)=P(B| C)=E(I_B|C)=E(Z|C)
=fracE(ZI_C)p(C)=fracE(I_BI_C)p(C)=fracE(I_Bcap C)p(C)=fracp(Bcap C)p(C)=fracp(Xleq acapX>3 )p(X>3 )$

now it is clear to continue.

for second topic:

$P(U<V)=E(I_U<V)=E(Z)=EE(Z|V)
=sum_t=1^n E(Z|V=t) p(V=t)=
sum_t=1^n E(I_U<V|V=t)p(V=t)=
sum_t=1^n E(I_U<t|V=t)p(V=t)=
sum_t=1^n P(U<t|V=t)p(V=t)$

note that by definition $E(Z|V)$ is a function of $V$

if you want generalize it if follow this step

$win R_w=uin 1,cdots , n,vin 1,cdots , n $

$P(U<V)=E(I_U<V)=EE(I_U<V|W=(V,U))=
EE(I_U<V|W)=sum_win R_w E(I_U<V|W=(V,U)=w) P(W=w)
=sum_v=1^n sum_u=1^n E(I_U<V|(V=v,U=u)) P(V=v,U=u)
=sum_v=1^n sum_u=1^n P(U<V|(V=v,U=u)) P(V=v,U=u)
=sum_v=1^n sum_u=1^n P(U<V|(V=v,U=u)) P(V=v|U=u)P(U=u)
$

But Note $P(U<V|U)$ is not generalize previous step.
$P(U<V|U)=P(A|U)=E(I_A|U)$ and
$E(I_A|U=u) = sum_v=1^n I_A P(V=v|U=u)=
sum_u<v,v=1^n P(V=v|U=u)=sum_v=u+1^n P(V=v|U=u)$

you should explain more.