# Risky Assets over Bernoulli Utility

My question is about 6.C.4. of Mas-Colell et.al.'s Microeconomic Theory book.

We have N risky assets with returns $z_n, n = (1,...,N)$ per dollar invested which are distributed over $F(z_1,...,z_n).$ All returns are non-negative with probability one. For an individual with a continuous, increasing, and concave Bernoulli utility function $u(\cdot)$ over $\mathbb{R_+}$, we define the utility function $U(\cdot)$ over $\mathbb{R}_+^N$, the set of nonnegative portfolios as:

$$U(\alpha_1,...,\alpha_N) = \int u(\alpha_1 z_1 + \cdots + \alpha_N z_N)dF(z_1,...,z_N)$$

We are asked to show for part c that $U(\cdot)$ is continuous.

The book's solution is a little difficult (particularly since I haven't really worked with measure theory before), but my teacher has said there is a more straightforward way of doing the question. I have questions about both of these points.

First, looking at the book's argument.

We create a sequence $(\alpha^m)_{m \in N} \to \alpha \in \mathbb{R}_+^N$. Then if $U$ is continuous, we should get $$(\alpha^m)_{m \in \mathbb{N}} \to \alpha \implies U(\alpha^m)_{m \in \mathbb{N}} \to U(\alpha)$$

$\exists \ \delta > 0$ s.t. $\alpha^m \leq (\delta,...,\delta) \ \forall \ m$

Since $U(\delta,...,\delta)$ is finite, $z \to u(\sum_n \delta z_n)$ is integrable.

Since $u(\cdot)$ monotone and returns are nonnegative with prob. zero:

$$u(\sum_n \alpha_n^m z_n) \leq u(\sum_n \delta z_n) \ \forall \ m, (z_1,...,z_N)$$

Since $u(\cdot)$ continuous,

$$u(\sum_n \alpha_n^m z_n) \to u(\sum_n \alpha_n z_n)$$ for almost all $(z_1,...,z_N)$

Here the book decides to apply Lebesgue's dominated convergence theorem. We are in some measure space $(S, \Sigma, \mu)$ which states that for some sequence of functions ${f_n}$, if it pointwise converges to some function $f$ and

$$\mid f_n(x) \mid \leq g(x) \ \forall n \quad \text{in index}, \ \forall x \in S$$

(recall we constructed $g(x)$ earlier)

Then $F$ is integrable and

$$\lim_{n \to \infty} \int_S \mid f_n - f \ \mid d \mu = 0$$

$$\implies \int_S \mid f_n - f \ \mid d \mu \to f d \mu$$

So

$$\int u(\sum_n \alpha_n^m x_n) dF(x_1,...,x_N) \to \int u(\sum_n \alpha_n x_n) dF(x_1,...,x_N)$$

$$\implies U(\alpha^m)_{m \in \mathbb{N}} \to U(\alpha)$$

I added in the direct definition of Lebesgue's dominated convergence for my own clarity, and I think I understand the proof, but my question for this part is what is the measure space? I know $S = \mathbb{R}_+^N$, and $\mu = F$, but what is $\Sigma$ supposed to be? I think it's supposed to be a $\sigma$ - algebra, but I have no idea how to construct/find it here. It is also very well possible that I am completely mistaken on what space I am working on, so there's that.

So then I tried to setup a proof for this question on my own. I also set up a sequence.

$(\alpha^m)_{m \in N} \to \alpha \in \mathbb{R}_+^N$. Then I suppose that there is some sort of jump discontinuity in $U$ and show that it leads to a contradiction.

So $(\alpha^m)_{m \in \mathbb{N}} \to \alpha$ and $U(\alpha^m)_{n \in \mathbb{N}} \not \to U(\alpha)$

My attempts so far to find a satisfactory proof have been fruitless. Any help would be appreciated, whether it's the proof itself or just an outline or some hints on what information to use from the original question to use.

Regarding your first question, the space on which you can apply Lebesgue's theorem is $\mathbb{R}_{+}^{N}$. The relevant $\sigma$-algebra is the Borel $\sigma$-algebra and the integration corresponds to the Lebesgue measure. Formally, with your notation, the function $f_m$ is defined on vectors $x=(x_1,\cdots,x_N)$ by \begin{equation*} f_m(x_1,\cdots,x_N) = u(\sum_{n}{\alpha_n^m x_n}) \end{equation*}
Regarding your second question, it seems to me that constructing a counter-example would be difficult and tedious. Proofs of continuity are rarely done that way. You might ask your teacher what kind of argument he has in mind. A possibility that you might wish to explore is to go back to the definition of continuity. Basically, you want to show that: \begin{equation*} \forall \alpha, \forall \epsilon>0, \exists \delta>0 \text{ such that } \|\alpha-\beta\|<\delta \Rightarrow \|U(\alpha)-U(\beta)\|<\epsilon \end{equation*} You can try to replace $U(\alpha)$ and $U(\beta)$ by their expression and manipulate the resulting integral. But this essentially boils down to proving Lebesgue's convergence theorem again : this solution is therefore not going to be simpler, but it might help you to understand what is behind this result.
You want to prove that $$U(\alpha) = \int_z u(\alpha\cdot z) dF(z)$$ is continuous, given that $$u(w)$$ is continuous?
By definition $$|U(\alpha)-U(\alpha')| = |\int_z u(\alpha\cdot z) - u(\alpha' \cdot z) dF(z) | <\int_z |u(\alpha\cdot z) - u(\alpha' \cdot z)| dF(z)$$. Take the supremum over $$z$$, to get $$\int_z |u(\alpha\cdot z) - u(\alpha' \cdot z)| dF(z) < \sup_{z} |u(\alpha\cdot z) - u(\alpha' \cdot z)| = |u(\alpha\cdot \xi) - u(\alpha' \cdot \xi)|$$ Now since $$u(w)$$ is continuous, take $$||\alpha-\alpha'||$$ sufficiently close that $$|\alpha\cdot \xi - \alpha' \cdot \xi| < \delta$$, where $$\delta$$ is sufficiently small that $$|w-w'|<\delta$$ implies $$|u(w) - u(w')| <\varepsilon$$. This implies that $$|U(\alpha)-U(\alpha')| < |u(\alpha\cdot \xi) - u(\alpha' \cdot \xi)| < \varepsilon.$$