a simple argument? This argument appeared in a proof in a paper (I am paraphrasing), and I am not sure why it is true. It should be rather simple.
Let $(\Omega,P)$ be some sample space. Let $X$ be a random variable between $0$ and $1$. Let $U \subseteq \Omega$ and let
$U^\prime = ${$ \omega \colon X(\omega) \ge E[X \mid U] - 1/m$}.
Then $P(U^\prime) \ge P(U)/m$.
Can't see exactly why this is true.
 A: We can assume that $U = \Omega$. So you want to show $$\Pr[X \geq E[X] - \alpha] \geq \alpha,$$ where $\alpha = 1/m$. If not, then $$E[X] < \alpha + (1-\alpha)(E[X] - \alpha) = (1-\alpha) E[X] + \alpha^2,$$ so that $\alpha E[X] < \alpha^2$ and $E[X] < \alpha$. But in that case, the inequality trivially holds.
EDIT: Some more details. Suppose that $\Pr[X > E[X] - \alpha] = p < \alpha$. Given $p$, the distribution maximizing $E[X]$ under this constraint is $$\Pr[X=E[X]-\alpha]=1-p,\quad \Pr[X=1]=p.$$ Its expectation is $$p + (1-p)(E[X] - \alpha)= E[X] - \alpha + p(1 - E[X] + \alpha).$$ Since $p < \alpha$ and $1 - E[X] + \alpha > 0$, we get that $$E[X] < E[X] - \alpha + \alpha(1 - E[X] + \alpha).$$ Rearranging, $$\alpha (1 + E[X]) < \alpha (1 + \alpha)$$ and so $E[X] < \alpha$, in which case the inequality trivially holds.
A: We want to show that 
$$
{\rm P}[X \ge {\rm E}(X|U) - 1/m] \ge {\rm P}(U)/m.
$$
This is trivially not true if $0 < m < 1$, for if $U = \Omega$, the right-hand side is greater than $1$. So, assume that $m \geq 1$, and ${\rm P}(U)>0$. Further, if ${\rm E}(X|U) - 1/m \leq 0$, then the inequality is trivial; hence we assume that
${\rm E}(X|U) - 1/m > 0$. Let $P_{X|U}$ denote the conditional distribution of $X$ given $U$, so that
$$
P_{X|U} {(A)} = {\rm P}(X \in A | U) = \frac{{{\rm P}([X \in A],U)}}{{{\rm P}(U)}}.
$$
Then,
$$
{\rm E}(X|U) = \int_{[0,1]} {xP_{X|U} ({\rm d}x)}. 
$$
Next, decompose $[0,1]$ into the union of the intervals $I_1 = \lbrace 0 \leq x < {\rm E}(X|U) - 1/m \rbrace$ and $I_2 = \lbrace {\rm E}(X|U) - 1/m \leq x \leq 1 \rbrace$. On the one hand,
$$
\int_{I_1 } {xP_{X|U} ({\rm d}x)}  \le  [{\rm E}(X|U) - 1/m]P_{X|U} (I_1 ) \le {\rm E}(X|U) - 1/m,
$$ 
and on the other hand, 
$$
\int_{I_2 } {xP_{X|U} ({\rm d}x)} \leq  \int_{I_2 } {P_{X|U} ({\rm d}x)} = P_{X|U} {(I_2)} \leq \frac{{{\rm P}(X \in I_2 )}}{{{\rm P}(U)}}.
$$
Combining it all, we get
$$
{\rm E}(X|U) \leq {\rm E}(X|U) - 1/m + {\rm P}(X \in I_2)/{\rm P}(U).
$$
That is ${\rm P}(X \in I_2) \geq {\rm P}(U)/m$, which is exactly what we want.
A: looks like chebyshev inequality.
