I am stuck on a question for quite sometime now, although in the text it is said to be "apparent". The problem goes as the following :

Let $X$ and $Y$ be locally compact Hausdorff spaces. Then $M(X)$ denote the space of all regular complex (hence bounded) measures on $X$, and $M^+(X)$ denote the space of all non-negative measures in $M(X)$. Same holds for $Y$.

We have a linear function $\phi : M(X) \rightarrow M(Y)$ given by $\mu \mapsto \mu'$ such that

(i) If $\mu \geq 0$ then $\mu' \geq 0$.

(ii) The restricted mapping $\phi|_{M^+(X)}$ from $M^+(X)$ to $M^+(Y)$ is continuous.

Let $p_x$ be the point-mass measure at $x \in X$. Also, suppose that $m$ is a non-negative measure on $X$ and $g$ is a lower semi-continuous function on $Y$.

Define a measure $m'$ on $Y$ by $m' := \int_X p_x' \ dm(x)$, and a function $g'$ on $X$ by $g'(x) := \int_Y g(y) \ dp_x'(y)$.

Now it is apparent from the definitions that $\int_Xg' \ dm = \int_Y g \ dm'$.

I need to show that $g'$ is lower semi-continuous on $X$.

What I already have is that :

(a) The number $M := \sup_{x \in X} ||p_x'||$ is finite.

(b) If $h$ is a bounded continuous function on $Y$, then $h'$ is continuous and $||h'||_\infty \leq M ||h||_\infty$.

(c) The function $x \mapsto p_x$ is a homeomorphism from $X$ onto a closed subset of $M^+(X)$.

Any kind of help/ comments will be Really appreciated ! :)

Please note that I have also asked this question in SE https://math.stackexchange.com/questions/528243/a-problem-concerning-measures-on-locally-compact-spaces , but did not find any answers there.

  • $\begingroup$ Also posted on math.SE. $\endgroup$ – Zev Chonoles Oct 16 '13 at 9:43
  • 2
    $\begingroup$ Hi, and welcome to MO! As Zev pointed out over at MSE, cross-posting is frowned upon, so it might be good to withdraw the question at one or the other site. (I'd have given it a little more time over there; it was asked just a few hours ago in the wee hours of the morning in North American time zones.) Generally, I think waiting at least two days at MSE before giving up there is appropriate. See also mathoverflow.net/help/on-topic for what sorts of questions are acceptable here. $\endgroup$ – Todd Trimble Oct 16 '13 at 11:34

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