There is another way I was just thinking of this vector valued integral, and several people (in particular, Paul Garrett) have already explained things to me in this way, but I was not able to understand what they were saying at the time. I will write what I have, and if it's wrong, hopefully someone will point out my error.

So we choose a maximal compact open subgroup $K$ of $G$, in good position relative to $P$, so that we have $G = PK$. For each $n \in N_w$, we choose $p_n \in P$ and $k_n \in K$ such that $w^{-1}n = p_nk_n$. Define $f_{\nu}(n) = q^{\langle \nu + \rho, H_M(p_n)\rangle}$. This is a well defined locally constant function on $N_w$. Also, for a given $f \in I(\nu,\pi)$, the map $n \mapsto f(k_n)$ is a well defined locally constant function $N_w \rightarrow V$, and we have

$$f(w^{-1}n) = f_{\nu}(n)f(k_n) \in V$$

Since $f|_K:K \rightarrow V$ is locally constant, $\{f(k_n) : n \in N_w \}$ is a finite set.

Now to make sense out of $\int\limits_{N_w} f(w^{-1}ng)dn$, we may replace $f$ by $R_g(f)$ and assume $g = 1$. Thus we want to make sense out of $\int\limits_{N_w} f(w^{-1}n)dn$. For $v \in V$, define

$$ A_v = \{ n \in N_w : f(k_n) = v\}$$

which is an open set in $N_w$, and is empty for almost all $v$. Naively, I'm going to write

$$\int\limits_{N_w} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)f(k_n)dn$$

$$ = \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$$ where the sum over $V$ is really just a finite sum. So the vector valued integral over $N_w$ can be defined to be $ \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$, provided each Lebesgue integral

$$\int\limits_{A_v}f_{\nu}(n)dn$$

converges. But each such Lebesgue integral converges if and only if

$$\sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)dn = \int\limits_{N_w} f_{\nu}(n)dn$$

converges. So the intertwining operator makes sense provided that $\int\limits_{N_w} |f_{\nu}(n)|dn < \infty$.

EDIT: This doesn't work, because $n \mapsto f(k_n)$ is not well defined.

There is another way I was just thinking of this vector valued integral, and several people (in particular, Paul Garrett) have already explained things to me in this way, but I was not able to understand what they were saying at the time. I will write what I have, and if it's wrong, hopefully someone will point out my error.

So we choose a maximal compact open subgroup $K$ of $G$, in good position relative to $P$, so that we have $G = PK$. For each $n \in N_w$, we choose $p_n \in P$ and $k_n \in K$ such that $w^{-1}n = p_nk_n$. Define $f_{\nu}(n) = q^{\langle \nu + \rho, H_M(p_n)\rangle}$. This is a well defined locally constant function on $N_w$. Also, for a given $f \in I(\nu,\pi)$, the map $n \mapsto f(k_n)$ is a well defined locally constant function $N_w \rightarrow V$, and we have

$$f(w^{-1}n) = f_{\nu}(n)f(k_n) \in V$$

Since $f|_K:K \rightarrow V$ is locally constant, $\{f(k_n) : n \in N_w \}$ is a finite set.

Now to make sense out of $\int\limits_{N_w} f(w^{-1}ng)dn$, we may replace $f$ by $R_g(f)$ and assume $g = 1$. Thus we want to make sense out of $\int\limits_{N_w} f(w^{-1}n)dn$. For $v \in V$, define

$$ A_v = \{ n \in N_w : f(k_n) = v\}$$

which is an open set in $N_w$, and is empty for almost all $v$. Naively, I'm going to write

$$\int\limits_{N_w} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)f(k_n)dn$$

$$ = \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$$ where the sum over $V$ is really just a finite sum. So the vector valued integral over $N_w$ can be defined to be $ \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$, provided each Lebesgue integral

$$\int\limits_{A_v}f_{\nu}(n)dn$$

converges. But each such Lebesgue integral converges if and only if

$$\sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)dn = \int\limits_{N_w} f_{\nu}(n)dn$$

converges. So the intertwining operator makes sense provided that $\int\limits_{N_w} |f_{\nu}(n)|dn < \infty$.

There is another way I was just thinking of this vector valued integral, and it's possible that some people (in particular, Paul Garrett) have already attempted to explain things to me in this way, but I was not able to understand what they were saying at the time. I will write what I have, and if it's wrong, hopefully someone will point out my error.

So we choose a maximal compact open subgroup $K$ of $G$, in good position relative to $P$, so that we have $G = PK$. For each $n \in N_w$, we choose $p_n \in P$ and $k_n \in K$ such that $w^{-1}n = p_nk_n$. Define $f_{\nu}(n) = q^{\langle \nu + \rho, H_M(p_n)\rangle}$. This is a well defined locally constant function on $N_w$. Also, for a given $f \in I(\nu,\pi)$, the map $n \mapsto f(k_n)$ is a well defined locally constant function $N_w \rightarrow V$, and we have

$$f(w^{-1}n) = f_{\nu}(n)f(k_n) \in V$$

Since $f|_K:K \rightarrow V$ is locally constant, $\{f(k_n) : n \in N_w \}$ is a finite set.

Now to make sense out of $\int\limits_{N_w} f(w^{-1}ng)dn$, we may replace $f$ by $R_g(f)$ and assume $g = 1$. Thus we want to make sense out of $\int\limits_{N_w} f(w^{-1}n)dn$. For $v \in V$, define

$$ A_v = \{ n \in N_w : f(k_n) = v\}$$

which is an open set in $N_w$, and is empty for almost all $v$. Naively, I'm going to write

$$\int\limits_{N_w} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)f(k_n)dn$$

$$ = \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$$ where the sum over $V$ is really just a finite sum. So the vector valued integral over $N_w$ can be defined to be $ \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$, provided each Lebesgue integral

$$\int\limits_{A_v}f_{\nu}(n)dn$$

converges. But each such Lebesgue integral converges if and only if

$$\sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)dn = \int\limits_{N_w} f_{\nu}(n)dn$$

converges. Thus the intertwining operator makes sense provided that $\int\limits_{N_w} |f_{\nu}(n)|dn < \infty$.

There is another way I was just thinking of this vector valued integral, and several people (in particular, Paul Garrett) have already explained things to me in this way, but I was not able to understand what they were saying at the time. I will write what I have, and if it's wrong, hopefully someone will point out my error.

So we choose a maximal compact open subgroup $K$ of $G$, in good position relative to $P$, so that we have $G = PK$. For each $n \in N_w$, we choose $p_n \in P$ and $k_n \in K$ such that $w^{-1}n = p_nk_n$. Define $f_{\nu}(n) = q^{\langle \nu + \rho, H_M(p_n)\rangle}$. This is a well defined locally constant function on $N_w$. Also, for a given $f \in I(\nu,\pi)$, the map $n \mapsto f(k_n)$ is a well defined locally constant function $N_w \rightarrow V$, and we have

$$f(w^{-1}n) = f_{\nu}(n)f(k_n) \in V$$

Since $f|_K:K \rightarrow V$ is locally constant, $\{f(k_n) : n \in N_w \}$ is a finite set.

Now to make sense out of $\int\limits_{N_w} f(w^{-1}ng)dn$, we may replace $f$ by $R_g(f)$ and assume $g = 1$. Thus we want to make sense out of $\int\limits_{N_w} f(w^{-1}n)dn$. For $v \in V$, define

$$ A_v = \{ n \in N_w : f(k_n) = v\}$$

which is an open set in $N_w$, and is empty for almost all $v$. Naively, I'm going to write

$$\int\limits_{N_w} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)f(k_n)dn$$

$$ = \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$$ where the sum over $V$ is really just a finite sum. So the vector valued integral over $N_w$ can be defined to be $ \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$, provided each Lebesgue integral

$$\int\limits_{A_v}f_{\nu}(n)dn$$

converges. But each such Lebesgue integral converges if and only if

$$\sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)dn = \int\limits_{N_w} f_{\nu}(n)dn$$

converges. So the intertwining operator makes sense provided that $\int\limits_{N_w} |f_{\nu}(n)|dn < \infty$.