A small math theorem that it took me an embarrassing long while to prove. Quite trivial actually.

Theorem Let m be a function from R⁺ to R⁺ that is decreasing and continuous on the right. Define f from R⁺ to R⁺ as follows:

f(t) := inf { λ | m(λ) <= t }

then (1) f is decreasing and (2) f is continuous on the right.

Proof: the first fact is trivial. First, by right continuity of m, we have that m(f(t)) <= t < t + d, therefore by monotonicity of m,

f(t) >= inf { λ | m(λ) <= t+d } = f(t+d)

To prove the second, note that by the fact that f is decreasing, it suffices to show that for all positive real number a that

f^-1( (a,inf) ) = an open interval with left endpoint 0

in particular, we need to show that the interval

I := { t | inf{λ | m(λ) <= t} > a }

is open on the right. Suppose τ is an element of I. We prove by contradiction. Then let us assume that for all ε > 0 we have f(τ+ε) <= a. Then for all ε there exists λ_ε <= a such that m(λ_ε) <= τ + ε. Consider a sequence ε_k tending toward 0 from the right. We construct an associated sequence λ_k. Since λ_k are bounded, up to a subsequence they converge to λ'. Furthermore, up to a subsequence, we can choose λ_k to be either non-increasing or non-decreasing. Suppose the former, then λ_k tends to λ' on the right, the by right continuity of m, we know that m(λ_k) tend to m(λ'), so m(λ') <= τ. But λ' <= a, contradicting the fact that f(τ) > a. Suppose the latter, then λ_k tends to λ' from the left, then we must have m(λ') <= m(λ_k) for all k, from which we also conclude that m(λ')<= τ and λ' <= a and arrive at the same contradiction. Therefore there must exist some ε > 0 such that f(τ+ε) < a, and hence I is open on the right.

Q.E.D.