After some helpful comments, I realized that I had to repost this question in a more systematic way.

On a complete probability space, let $\mathcal{H}_0$ denote the Hilbert space of square integrable random variables with zero mean. A stochastic process $X$ is called a second order process if $\mathbf{E}X(t)^2 < \infty$ and $\mathbf{E}X(t) = 0$, all $t \in [0,T]$. Such a process can be regarded as a map $[0,T] \rightarrow \mathcal{H}_0$. It is called q.m. continuous if this map is continuous, i.e. $X(s) \rightarrow X(t)$ in quadratic mean as $s \rightarrow t$. One can show that each q.m. continuous process has a measurable version.

Let $X$ be a q.m. continuous second order process. We want to compute the integral $\int_0^T X(s) \mathrm{d} s$. There are two ways.

Bochner integral. Clearly, $X$ considered as a continuous map $[0,T] \rightarrow \mathcal{H}_0$ is Bochner integrable. We denote its Bochner integral by \begin{equation} \text{(B-)}\int_0^T X(s) \mathrm{d}s. \end{equation}

Lebesgue integral. We may assume that $X$ considered as a map $[0,T] \times \Omega \rightarrow \mathrm{R}$ is measurable. Thus, for fixed $\omega$, the integral $\int_0^T X(s,\omega) \mathrm{d} s$ exists as a Lebesgue integral, and we denote the random variable constructed in this way by \begin{equation} \text{(L-)}\int_0^T X(s) \mathrm{d}s. \end{equation}

Question. Do we have \begin{equation} \text{(B-)}\int_0^T X(s) \mathrm{d}s = \text{(L-)}\int_0^T X(s) \mathrm{d}s \quad \text{a.s.?} \end{equation}

Ideas. Let $\lbrace t^n = t_0^n, \ldots t_{k_n}^n \rbrace$ be a sequence of partitions of $[0,T]$ with mesh going to zero. Define the simple functions \begin{equation} \xi_n = X(t_0^n)1[t_0^n,t_1^n] + \sum_{i=1}^{k_n-1} X(t_i^n) 1[t_i^n, t_{i+1}^n). \end{equation} Then one can show that for almost every $t$, we have $\xi_n(t) \rightarrow X(t)$ in $\mathcal{H}_0$, and \begin{equation} \int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d}s \quad \text{in $\mathcal{H}_0$}, \end{equation} where the integral on the left is defined in the obvious way (we omit (B-)) (to show this, one uses the fact that the covariance function $r(s,t) = \mathbf{E}X(s)X(t)$ of a q.m. continuous process is continuous). After switching to a subsequence if necessary, we may assume that \begin{equation} \int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d}s \quad \text{$\mathbf{P}$-a.s.}, \end{equation} Now, we would like to have that also \begin{equation} \int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(L-)}\int_0^T X(s) \mathrm{d}s \quad \text{$\mathbf{P}$-a.s.}, \end{equation} But this is tricky. The sums on the left hand side are Riemann sums, i.e. \begin{equation} \int_0^T \xi_n(s) \mathrm{d}s = \sum_{i=0}^{k_n-1} X(t_i^n)(t_{i+1}^n - t_i^n ). \end{equation} So if we knew that the paths of $X$ are a.s. Riemann integrable, we would be done. But this is not clear. I also tried to use some approximation arguments, but couldn't do it. It seems like one needs to deduce some kind of path regularity of $X$ from the assumption of q.m. continuity, but I don't know any results in this direction.

  • $\begingroup$ Perhaps I'm missing something, but it seems that the answer to your question is "yes" simply because the projection of any $Y \in \mathcal{H}_0$ onto $\mathcal{L}(X,t)$ is exactly $\mathbb{E}[Y|\mathcal{F}_t]$, where $\mathcal{F}_t = \sigma(X_s : s \le t)$. And of course $\int_0^tX_sds$ is $\mathcal{F}_t$-measurable. $\endgroup$
    – Dan
    Oct 28, 2012 at 14:45
  • $\begingroup$ The assertion that the orthogonal projection equals the conditional expectation is only true for Gaussian processes. In general, the conditional expectation is the orthogonal projection onto $L_2(\Omega,\mathcal{X}(t),\mathbf{P})$, $\mathcal{X}$ the natrual filtration of $X$, which is a larger subspace than $\mathcal{L}(X,t)$. So in general, these two are not equal. $\endgroup$
    – Hauke L.
    Oct 28, 2012 at 15:15
  • $\begingroup$ *orthogonal projection onto $\mathcal{L}(X,t)$ i should say (1st sentence) $\endgroup$
    – Hauke L.
    Oct 28, 2012 at 15:30
  • $\begingroup$ Of course, of course, hence the "linear". $X :[0,T] \rightarrow \mathcal{H}$ is certainly Bochner integrable, so if by $\int_0^tX(s)ds$ you mean the Bochner integral, then the answer is yes. But I suppose the point is that the Bochner integral may disagree with the pathwise Lebesgue or Riemann integral. $\endgroup$
    – Dan
    Oct 28, 2012 at 17:22
  • $\begingroup$ I did not know the Bochner integral before. And yes, the point is whether this integral is a.s. equal to the Lebesgue integral defined path by path. This does not seem obvious to me. $\endgroup$
    – Hauke L.
    Oct 28, 2012 at 18:05

1 Answer 1


Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question.

For a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $X\colon[0,T]\to L^p(\mathbb{P})$ ($1\le p\le\infty$) be Bochner integrable w.r.t the Lebesgue measure on $[0,T]$, and also jointly measurable as a map $(t,\omega)\mapsto X(t)(\omega)$ from $[0,T]\times\Omega$ to $\mathbb{R}$. Then, the Bocher integral $\int_0^T X(t)\,dt$ agrees with the pathwise Lebesgue integral $\int_0^TX(t)(\omega)\,dt$ for almost every $\omega$.

First, this statement clearly holds for simple functions, which are finite linear combinations of terms of the form $X(t)(\omega)=1_{\lbrace t\in A\rbrace}1_{\lbrace\omega\in B\rbrace}$, for $A$ a Borel subset of $[0,T]$ and $B$ in $\mathcal{F}$. Now, by definition, if $X$ is Bochner integrable then, for each $n\ge1$, there is a simple $\xi_n$ such that $$ \int_0^T\lVert X(t)-\xi_n(t)\rVert_p\,dt\le2^{-n}. $$ The Bochner integral is given by $$ \int_0^T\xi_n(t)\,dt \rightarrow\text{(B-)}\int_0^T X(t)\,dt. $$ Here the limit is taken in the $L^p$ norm and, hence, also holds for convergence in probability.

Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs. $$ \begin{align} \mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt\right] &=\sum_{n=1}^\infty\int_0^T\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\,dt\cr &\le\sum_{n=1}^\infty\int_0^T\left\lVert X(t)-\xi_n(t)\right\rVert_p\,dt\cr &\le\sum_{n=1}^\infty2^{-n}=1 < \infty. \end{align} $$ In particular, $$ \int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt < \infty $$ with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies, $$ \int_0^T\xi_n(t)\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\,dt. $$ This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.

  • 1
    $\begingroup$ Now i understand what you meant in your previous comment. Thanks a lot. $\endgroup$
    – Hauke L.
    Oct 30, 2012 at 11:51
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    $\begingroup$ You can also say that Bochner integral = pathwise Lebesgue integral in rather more generality. I'll update the answer later to incorporate this. $\endgroup$ Oct 30, 2012 at 12:15
  • $\begingroup$ I updated my answer now. $\endgroup$ Nov 8, 2012 at 1:22
  • $\begingroup$ Hi, George: Would you mind taking a look at my question mathoverflow.net/q/301507/32660 regarding stochastic process convergence? Thank you. $\endgroup$
    – Hans
    May 29, 2018 at 22:16

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